Measures of dispersion

1. Revision of range


ID is: 3151 Seed is: 5565

Working with the range of a data set

Determine the minimum value in a data set if the maximum value in the data set is 43 and the range of the data is 0.

Answer: The minimum value is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Remember that the range is the difference between the largest and smallest values in a data set.
STEP: Use the range to find the answer
[−1 point ⇒ 0 / 1 points left]

The range of a data set tells us how spread out the data are: it is the difference between the largest and smallest values. In this question, we do not know what the data values are, but we know that the range of the data is 0. We also know that the maximum value is 43.

We can solve this question using the equation for the range:

range = maximum - minimum

However, for this solution let's think through the information that we have. In this case, the range is zero. That means there is no spread in the data: all the values must be equal. In fact, the data must be something like this:

{43;43;43}

We do not know how many values there are in the set (there is no information about that). But if the range is zero then there is no separation between the largest and smallest values. All the values must be the same.

The minimum value is 43.


Submit your answer as:

ID is: 3151 Seed is: 8043

Working with the range of a data set

Consider a data set with a maximum value of 22. What is the minimum value in the data set if the range of the data is 0?

Answer: The minimum value is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Remember that the range is the difference between the largest and smallest values in a data set.
STEP: Use the range to find the answer
[−1 point ⇒ 0 / 1 points left]

The range of a data set tells us how spread out the data are: it is the difference between the largest and smallest values. In this question, we do not know what the data values are, but we know that the range of the data is 0. We also know that the maximum value is 22.

We can solve this question using the equation for the range:

range = maximum - minimum

However, for this solution let's think through the information that we have. In this case, the range is zero. That means there is no spread in the data: all the values must be equal. In fact, the data must be something like this:

{22;22;22;22}

We do not know how many values there are in the set (there is no information about that). But if the range is zero then there is no separation between the largest and smallest values. All the values must be the same.

The minimum value is 22.


Submit your answer as:

ID is: 3151 Seed is: 2053

Working with the range of a data set

Consider a data set with a maximum value of 33. What is the minimum value in the data set if the range of the data is 17?

Answer: The minimum value is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Remember that the range is the difference between the largest and smallest values in a data set.
STEP: Use the range to find the answer
[−1 point ⇒ 0 / 1 points left]

The range of a data set tells us how spread out the data are: it is the difference between the largest and smallest values. In this question, we do not know what the data values are, but we know that the range of the data is 17. We also know that the maximum value is 33.

We can solve this question using the equation for the range:

range = maximum - minimum

However, for this solution let's think through the information that we have. The maximum value in the data set is 33 and the range is 17. So the minimum value is 17 below 33.

minimum = maximum - range=3317=16

The minimum value is 16.


Submit your answer as:

ID is: 3154 Seed is: 1391

Finding the range by solving an equation

Here is some information about a certain data set:

  1. The minimum value is 6a.
  2. The maximum value is 9.
  3. The range of the data is equal to 3(a5).

Use this information to determine the range of data set.

Answer: The range of the data is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Write an equation which relates the range to the maximum and minimum values given. Then solve for a.


STEP: Write an equation with the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the range of a data set, where we know the minimum, the maximum and the range in terms of a. We can solve this question using the definition of the range.

range = maximum - minimum

Substituting in the expressions from the question, we get:

3(a5)=(9)(6a)

STEP: Solve the equation
[−1 point ⇒ 1 / 3 points left]

Fantastic: now we have an equation we can solve. Start by expanding the expression on the left.There is also a double negative on the right hand side changes to addition.

3a+15=9+6a3a+15=6a+99a=6a=23

STEP: Calculate the range
[−1 point ⇒ 0 / 3 points left]

Now we know that a=23. But we are not finished yet. The question asked us to find the range of the data set. We can use the value of a to do that. Just substitute a=23 into the expression for the range. Use the expanded expression for the range, which we found above (this saves a little time).

range=3a+15=3(23)+15=13

The range of the data set is 13.


Submit your answer as:

ID is: 3154 Seed is: 3712

Finding the range by solving an equation

Here is some information about a certain data set:

  1. The minimum value is 6n+17.
  2. The maximum value is 34.
  3. The range of the data is equal to 8(n3).

Use this information to determine the range of data set.

Answer: The range of the data is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Write an equation which relates the range to the maximum and minimum values given. Then solve for n.


STEP: Write an equation with the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the range of a data set, where we know the minimum, the maximum and the range in terms of n. We can solve this question using the definition of the range.

range = maximum - minimum

Substituting in the expressions from the question, we get:

8(n3)=(34)(6n+17)

STEP: Solve the equation
[−1 point ⇒ 1 / 3 points left]

Fantastic: now we have an equation we can solve. Start by expanding the expression on the left. Notice that on the right hand side we also must distribute the negative sign into the binomial 6n+17.

8n+24=34(6n+17)8n+24=34+6n178n+24=6n+1714n=7n=12

STEP: Calculate the range
[−1 point ⇒ 0 / 3 points left]

Now we know that n=12. But we are not finished yet. The question asked us to find the range of the data set. We can use the value of n to do that. Just substitute n=12 into the expression for the range. Use the expanded expression for the range, which we found above (this saves a little time).

range=8n+24=8(12)+24=20

The range of the data set is 20.


Submit your answer as:

ID is: 3154 Seed is: 4141

Finding the range by solving an equation

Here is some information about a certain data set:

  1. The minimum value is 3a15.
  2. The maximum value is 9a26.
  3. The range of the data is equal to 9a7.

Use this information to determine the range of data set.

Answer: The range of the data is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Write an equation which relates the range to the maximum and minimum values given. Then solve for a.


STEP: Write an equation with the given information
[−1 point ⇒ 2 / 3 points left]

The question asks us to find the range of a data set, where we know the minimum, the maximum and the range in terms of a. We can solve this question using the definition of the range.

range = maximum - minimum

Substituting in the expressions from the question, we get:

9a7=(9a26)(3a15)

STEP: Solve the equation
[−1 point ⇒ 1 / 3 points left]

Fantastic: now we have an equation we can solve. Notice that on the right hand side we must distribute the negative sign into the binomial 3a15.

9a7=9a26(3a15)9a7=9a26+3a+159a7=12a113a=4a=43

STEP: Calculate the range
[−1 point ⇒ 0 / 3 points left]

Now we know that a=43. But we are not finished yet. The question asked us to find the range of the data set. We can use the value of a to do that. Just substitute a=43 into the expression for the range.

range=9a7=9(43)7=5

The range of the data set is 5.


Submit your answer as:

ID is: 4351 Seed is: 9011

Median, interquartile range, box & whisker diagram

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the numbers of devices in use by 19 schools for Siyavula Practice.

1199890155181151681341711079561131809614410155117
  1. Determine the median and the interquartile range of the data.
    Answer:
    • The median is devices.
    • The interquartile range is devices.
    numeric
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    • The median is the middle value in an ordered list.
    • The upper quartile is the median of the upper half of the data values.
    • The lower quartile is the median of the lower half of the data values.
    • The interquartile range is the difference between the upper and lower quartiles.

    STEP: Determine the median of the data set
    [−1 point ⇒ 2 / 3 points left]
    The median is the value exactly in the middle when a data set is ordered from minimum to maximum. Hence, a data set must first be ordered before we can identify the median.

    The data values:

    1199890155181151681341711079561131809614410155117

    The ordered data values:

    5561688090959698101107117119131134144151155171181

    The data set has 19 data values. The median is the tenth value in the ordered list, which is 107 devices.


    STEP: Determine the interquartile range
    [−2 points ⇒ 0 / 3 points left]
    • The median is the ‘middle value’ of the whole data set, referred to as Q2.
    • The lower quartile is the median of the lower half of the data values, referred to as Q1.
    • The upper quartile is the median of the upper half of the data values, referred to as Q3.
    • The interquartile range is the difference between the upper and lower quartiles, referred to as IQR. The formula is: IQR=Q3Q1.

    We already calculated the median (the ‘middle value’) of the whole data set. Now, we must calculate the median of the lower half and the median of the upper half.

    The ordered data values:

    5561688090959698101107117119131134144151155171181

    Calculate the lower and upper quartiles:

    Q1=T5=90 devices
    Q3=T15=144 devices

    Calculate the interquartile range:

    IQR=Q3Q1=14490=54 devices

    Submit your answer as: and
  2. On a piece of paper, draw a box and whisker diagram for the given data set:

    5561688090959698101107117119131134144151155171181
    NOTE: You should always draw a box and whisker diagram on a scaled number line.

    Use your diagram to give the exact values of the reference points shown below. (This sketch is not to scale.)

    Answer:

    The value at:

    • a is devices.
    • b is devices.
    • c is devices.
    • d is devices.
    • e is devices.
    numeric
    one-of
    type(numeric.abserror(0.01))
    numeric
    one-of
    type(numeric.abserror(0.01))
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    When you draw a box and whisker diagram, you mark the following values on the number line:

    • The minimum value
    • The lower quartile
    • The median of the whole data set
    • The upper quartile
    • The maximum value

    STEP: Draw box and whisker diagram
    [−3 points ⇒ 0 / 3 points left]

    A box and whisker diagram is used to show information about the range, the median, and the quartiles of a data set. It is constructed in the following way:

    • A number line is drawn showing the range of the data set.
    • A box, representing the interquartile range, is drawn on the number line.
    • The left edge of the box shows the lower quartile.
    • The right edge of the box shows the upper quartile.
    • A line from the minimum value to the box (called the left whisker) and a line from the box to the maximum value (called the right whisker) are also drawn.
    • The median of the whole data set is shown as a vertical line inside the box.

    The labels you need for your box and whisker diagram are:

    • a shows the lowest value in the data set, the minimum value, which is 55 devices.
    • b shows the left edge of the box, the lower quartile of the data set, which is 90 devices.
    • c shows the line inside the box, the median of the whole data set, which is 107 devices.
    • d shows the right edge of the box, the upper quartile of the data set, which is 144 devices.
    • e shows the highest value in the data set, the maximum value, which is 181 devices.

    When drawn to scale, the box and whisker diagram looks like this:

    On your diagram, you must label all five number values, but you do not need to label the words the box, min, max, median, the left 'whisker', the right 'whisker', or the quartile labels Q1, Q2, and Q3.

    Submit your answer as: andandandand

ID is: 4351 Seed is: 4795

Median, interquartile range, box & whisker diagram

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the numbers of books donated by 11 classes for a literacy programme.

743740345565520317142
  1. Determine the median and the interquartile range of the data.
    Answer:
    • The median is books.
    • The interquartile range is books.
    numeric
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    • The median is the middle value in an ordered list.
    • The upper quartile is the median of the upper half of the data values.
    • The lower quartile is the median of the lower half of the data values.
    • The interquartile range is the difference between the upper and lower quartiles.

    STEP: Determine the median of the data set
    [−1 point ⇒ 2 / 3 points left]
    The median is the value exactly in the middle when a data set is ordered from minimum to maximum. Hence, a data set must first be ordered before we can identify the median.

    The data values:

    743740345565520317142

    The ordered data values:

    520313437404255567174

    The data set has 11 data values. The median is the sixth value in the ordered list, which is 40 books.


    STEP: Determine the interquartile range
    [−2 points ⇒ 0 / 3 points left]
    • The median is the ‘middle value’ of the whole data set, referred to as Q2.
    • The lower quartile is the median of the lower half of the data values, referred to as Q1.
    • The upper quartile is the median of the upper half of the data values, referred to as Q3.
    • The interquartile range is the difference between the upper and lower quartiles, referred to as IQR. The formula is: IQR=Q3Q1.

    We already calculated the median (the ‘middle value’) of the whole data set. Now, we must calculate the median of the lower half and the median of the upper half.

    The ordered data values:

    520313437404255567174

    Calculate the lower and upper quartiles:

    Q1=T3=31 books
    Q3=T9=56 books

    Calculate the interquartile range:

    IQR=Q3Q1=5631=25 books

    Submit your answer as: and
  2. On a piece of paper, draw a box and whisker diagram for the given data set:

    520313437404255567174
    NOTE: You should always draw a box and whisker diagram on a scaled number line.

    Use your diagram to give the exact values of the reference points shown below. (This sketch is not to scale.)

    Answer:

    The value at:

    • a is books.
    • b is books.
    • c is books.
    • d is books.
    • e is books.
    numeric
    one-of
    type(numeric.abserror(0.01))
    numeric
    one-of
    type(numeric.abserror(0.01))
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    When you draw a box and whisker diagram, you mark the following values on the number line:

    • The minimum value
    • The lower quartile
    • The median of the whole data set
    • The upper quartile
    • The maximum value

    STEP: Draw box and whisker diagram
    [−3 points ⇒ 0 / 3 points left]

    A box and whisker diagram is used to show information about the range, the median, and the quartiles of a data set. It is constructed in the following way:

    • A number line is drawn showing the range of the data set.
    • A box, representing the interquartile range, is drawn on the number line.
    • The left edge of the box shows the lower quartile.
    • The right edge of the box shows the upper quartile.
    • A line from the minimum value to the box (called the left whisker) and a line from the box to the maximum value (called the right whisker) are also drawn.
    • The median of the whole data set is shown as a vertical line inside the box.

    The labels you need for your box and whisker diagram are:

    • a shows the lowest value in the data set, the minimum value, which is 5 books.
    • b shows the left edge of the box, the lower quartile of the data set, which is 31 books.
    • c shows the line inside the box, the median of the whole data set, which is 40 books.
    • d shows the right edge of the box, the upper quartile of the data set, which is 56 books.
    • e shows the highest value in the data set, the maximum value, which is 74 books.

    When drawn to scale, the box and whisker diagram looks like this:

    On your diagram, you must label all five number values, but you do not need to label the words the box, min, max, median, the left 'whisker', the right 'whisker', or the quartile labels Q1, Q2, and Q3.

    Submit your answer as: andandandand

ID is: 4351 Seed is: 7970

Median, interquartile range, box & whisker diagram

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the numbers of ecobricks made by 25 classes during Enviro Week.

69235510318649483818624565321770434735394028535
  1. Determine the median and the interquartile range of the data.
    Answer:
    • The median is ecobricks.
    • The interquartile range is ecobricks.
    numeric
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    • The median is the middle value in an ordered list.
    • The upper quartile is the median of the upper half of the data values.
    • The lower quartile is the median of the lower half of the data values.
    • The interquartile range is the difference between the upper and lower quartiles.

    STEP: Determine the median of the data set
    [−1 point ⇒ 2 / 3 points left]
    The median is the value exactly in the middle when a data set is ordered from minimum to maximum. Hence, a data set must first be ordered before we can identify the median.

    The data values:

    69235510318649483818624565321770434735394028535

    The ordered data values:

    58910171823283132353839404345474853556264656970

    The data set has 25 data values. The median is the 13th value in the ordered list, which is 39 ecobricks.


    STEP: Determine the interquartile range
    [−2 points ⇒ 0 / 3 points left]
    • The median is the ‘middle value’ of the whole data set, referred to as Q2.
    • The lower quartile is the median of the lower half of the data values, referred to as Q1.
    • The upper quartile is the median of the upper half of the data values, referred to as Q3.
    • The interquartile range is the difference between the upper and lower quartiles, referred to as IQR. The formula is: IQR=Q3Q1.

    We already calculated the median (the ‘middle value’) of the whole data set. Now, we must calculate the median of the lower half and the median of the upper half.

    The ordered data values:

    58910171823283132353839404345474853556264656970

    Calculate the lower and upper quartiles:

    Q1=T6+T72=18+232=20.5
    Q3=T19+T202=53+552=54

    Calculate the interquartile range:

    IQR=Q3Q1=5420.5=33.5 ecobricks

    Submit your answer as: and
  2. On a piece of paper, draw a box and whisker diagram for the given data set:

    58910171823283132353839404345474853556264656970
    NOTE: You should always draw a box and whisker diagram on a scaled number line.

    Use your diagram to give the exact values of the reference points shown below. (This sketch is not to scale.)

    Answer:

    The value at:

    • a is ecobricks.
    • b is ecobricks.
    • c is ecobricks.
    • d is ecobricks.
    • e is ecobricks.
    numeric
    one-of
    type(numeric.abserror(0.01))
    numeric
    one-of
    type(numeric.abserror(0.01))
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    When you draw a box and whisker diagram, you mark the following values on the number line:

    • The minimum value
    • The lower quartile
    • The median of the whole data set
    • The upper quartile
    • The maximum value

    STEP: Draw box and whisker diagram
    [−3 points ⇒ 0 / 3 points left]

    A box and whisker diagram is used to show information about the range, the median, and the quartiles of a data set. It is constructed in the following way:

    • A number line is drawn showing the range of the data set.
    • A box, representing the interquartile range, is drawn on the number line.
    • The left edge of the box shows the lower quartile.
    • The right edge of the box shows the upper quartile.
    • A line from the minimum value to the box (called the left whisker) and a line from the box to the maximum value (called the right whisker) are also drawn.
    • The median of the whole data set is shown as a vertical line inside the box.

    The labels you need for your box and whisker diagram are:

    • a shows the lowest value in the data set, the minimum value, which is 5 ecobricks.
    • b shows the left edge of the box, the lower quartile of the data set, which is 20.5 ecobricks.
    • c shows the line inside the box, the median of the whole data set, which is 39 ecobricks.
    • d shows the right edge of the box, the upper quartile of the data set, which is 54 ecobricks.
    • e shows the highest value in the data set, the maximum value, which is 70 ecobricks.

    When drawn to scale, the box and whisker diagram looks like this:

    On your diagram, you must label all five number values, but you do not need to label the words the box, min, max, median, the left 'whisker', the right 'whisker', or the quartile labels Q1, Q2, and Q3.

    Submit your answer as: andandandand

ID is: 3153 Seed is: 7597

Percentile and quartiles

Percentile is a value which tells us how far along through an ordered data set a certain value sits. Percentile is related to minimum and maximum values in a data set and to the quartiles.

From the dropdown menu below, choose the value which corresponds to the 25th percentile.

Answer: The 25th percentile corresponds to the .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Starting at the minimum value and working up to the maximum value, percentile shows where a number sits in the data set. For example, the 50th percentile is halfway from the minimum value to the maximum value.
STEP: Use the definition of the quartiles
[−1 point ⇒ 0 / 1 points left]

Quartiles break a data set up into four groups. We get those four groups by 'breaking' the data set when the percentile is a multiple of 25.

Percentile Position in the data Value in the data set
0th beginning minimum value
25th 1 quarter of the way first quartile
50th 2 quarters of the way median (second quartile)
75th 3 quarters of the way third quartile
100th upper limit of the data maximum value

For this question we have the 25th percentile. This corresponds to the first quartile.

The correct choice from the dropdown menu is: first quartile.


Submit your answer as:

ID is: 3153 Seed is: 1673

Percentile and quartiles

Percentile is a value which tells us how far along through an ordered data set a certain value sits. Percentile is related to minimum and maximum values in a data set and to the quartiles.

From the dropdown menu below, choose the value which corresponds to the 25th percentile.

Answer: The 25th percentile corresponds to the .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Starting at the minimum value and working up to the maximum value, percentile shows where a number sits in the data set. For example, the 50th percentile is halfway from the minimum value to the maximum value.
STEP: Use the definition of the quartiles
[−1 point ⇒ 0 / 1 points left]

Quartiles break a data set up into four groups. We get those four groups by 'breaking' the data set when the percentile is a multiple of 25.

Percentile Position in the data Value in the data set
0th beginning minimum value
25th 1 quarter of the way first quartile
50th 2 quarters of the way median (second quartile)
75th 3 quarters of the way third quartile
100th upper limit of the data maximum value

For this question we have the 25th percentile. This corresponds to the first quartile.

The correct choice from the dropdown menu is: first quartile.


Submit your answer as:

ID is: 3153 Seed is: 6233

Percentile and quartiles

Percentile is a value which tells us how far along through an ordered data set a certain value sits. Percentile is related to minimum and maximum values in a data set and to the quartiles.

From the dropdown menu below, choose the value which corresponds to the 17th percentile.

Answer: The 17th percentile corresponds to the .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Starting at the minimum value and working up to the maximum value, percentile shows where a number sits in the data set. For example, the 50th percentile is halfway from the minimum value to the maximum value.
STEP: Use the definition of the quartiles
[−1 point ⇒ 0 / 1 points left]

Quartiles break a data set up into four groups. We get those four groups by 'breaking' the data set when the percentile is a multiple of 25.

Percentile Position in the data Value in the data set
0th beginning minimum value
25th 1 quarter of the way first quartile
50th 2 quarters of the way median (second quartile)
75th 3 quarters of the way third quartile
100th upper limit of the data maximum value

For this question we have the 17th percentile. This does not correspond to any of the quartile boundaries, which are all multiples of 25.

The correct choice from the dropdown menu is: none of the above.


Submit your answer as:

ID is: 3185 Seed is: 3455

Percentile and rank in a data set

The data set below shows a total of 14 values.

{a;c;g;i;j;k;l;m;o;r;s;u;v;w}

Note that the data are in alphabetical order.

  1. Find the rank of the data value at the tenth percentile.

    INSTRUCTION: Do not round off your answer.
    Answer: The rank of the tenth percentile is: .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the following equation to convert the percentile given in the question to a rank:

    r=p100(n1)+1

    STEP: Identify the values needed for the solution
    [−1 point ⇒ 1 / 2 points left]

    We have a data set with 14 items in it, and the question asks us to determine the rank of the tenth percentile in the data set. For this we use the formula:

    r=p100(n1)+1

    where:

    • p is the percentile value
    • n is the number of values in the data set
    • r is the rank (position) of the pth percentile in the data set

    In this case, we have the values:

    p=10n=14

    STEP: Substitute into the formula and evaluate the answer
    [−1 point ⇒ 0 / 2 points left]

    Now substitute the values p and n into the formula and evaluate to determine r. As always, we must follow the order of operations. That means starting with the subtraction in the brackets.

    r=p100(n1)+1=(10)100×((14)1)+1=10100×(13)+1=1.3+1=2.3

    The rank of the tenth percentile in the data set is 2.3.


    Submit your answer as:
  2. Identify the value in the data set which is at or closest to the tenth percentile. If there is more than one data value closest to the tenth percentile, give any one answer.

    Answer: The value at or closest to the tenth percentile is: .
    one-of
    type(string)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the rank value from Question 1 to find the value in the data set closest to the tenth percentile.


    STEP: Use the rank of the tenth percentile to find the correct value in the data set
    [−1 point ⇒ 0 / 1 points left]

    Now we need to find the value in the data set which corresponds to the tenth percentile. To do this, we use the rank we calculated in Question 1. Remember that the rank tells us the position of a value in an ordered list. In this case we want to know which value has a rank of 2.3.

    In truth, the rank of a value must be an integer (because it represents a position). So we need to find the closest position value to 2.3.

    The rank value 2.3 is closest to second position in the data set. So the tenth percentile corresponds to c.

    The value at or closest to the tenth percentile is c.


    Submit your answer as:

ID is: 3185 Seed is: 6814

Percentile and rank in a data set

The data set below shows a total of 7 values.

{20;16;8;3;2;14;25}

Note that the data are in numeric order.

  1. Find the rank of the data value at the 75th percentile.

    INSTRUCTION: Do not round off your answer.
    Answer: The rank of the 75th percentile is: .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the following equation to convert the percentile given in the question to a rank:

    r=p100(n1)+1

    STEP: Identify the values needed for the solution
    [−1 point ⇒ 1 / 2 points left]

    We have a data set with 7 items in it, and the question asks us to determine the rank of the 75th percentile in the data set. For this we use the formula:

    r=p100(n1)+1

    where:

    • p is the percentile value
    • n is the number of values in the data set
    • r is the rank (position) of the pth percentile in the data set

    In this case, we have the values:

    p=75n=7

    STEP: Substitute into the formula and evaluate the answer
    [−1 point ⇒ 0 / 2 points left]

    Now substitute the values p and n into the formula and evaluate to determine r. As always, we must follow the order of operations. That means starting with the subtraction in the brackets.

    r=p100(n1)+1=(75)100×((7)1)+1=75100×(6)+1=4.5+1=5.5

    The rank of the 75th percentile in the data set is 5.5.


    Submit your answer as:
  2. Figure out the value in the data set which is at or closest to the 75th percentile. If there is more than one data value closest to the 75th percentile, give any one answer.

    Answer: The value at or closest to the 75th percentile is: .
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the rank value from Question 1 to find the value in the data set closest to the 75th percentile.


    STEP: Use the rank of the 75th percentile to find the correct value in the data set
    [−1 point ⇒ 0 / 1 points left]

    Now we need to find the value in the data set which corresponds to the 75th percentile. To do this, we use the rank we calculated in Question 1. Remember that the rank tells us the position of a value in an ordered list. In this case we want to know which value has a rank of 5.5.

    In truth, the rank of a value must be an integer (because it represents a position). So we need to find the closest position value to 5.5.

    The rank value 5.5 is exactly between the fifth and sixth positions. So the 75th percentile corresponds equally well to both 2 and other 14.

    Either of these answers is acceptable for the value closest to the 75th percentile: 2 or 14.


    Submit your answer as:

ID is: 3185 Seed is: 9389

Percentile and rank in a data set

The data set below shows a total of 12 values.

{26;21;20;8;5;1;6;7;8;9;13;18}

Note that the data are in numeric order.

  1. Find the rank of the data value at the 49th percentile.

    INSTRUCTION: Do not round off your answer.
    Answer: The rank of the 49th percentile is: .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the following equation to convert the percentile given in the question to a rank:

    r=p100(n1)+1

    STEP: Identify the values needed for the solution
    [−1 point ⇒ 1 / 2 points left]

    We have a data set with 12 items in it, and the question asks us to determine the rank of the 49th percentile in the data set. For this we use the formula:

    r=p100(n1)+1

    where:

    • p is the percentile value
    • n is the number of values in the data set
    • r is the rank (position) of the pth percentile in the data set

    In this case, we have the values:

    p=49n=12

    STEP: Substitute into the formula and evaluate the answer
    [−1 point ⇒ 0 / 2 points left]

    Now substitute the values p and n into the formula and evaluate to determine r. As always, we must follow the order of operations. That means starting with the subtraction in the brackets.

    r=p100(n1)+1=(49)100×((12)1)+1=49100×(11)+1=5.39+1=6.39

    The rank of the 49th percentile in the data set is 6.39.


    Submit your answer as:
  2. Determine the value in the data set which is at or closest to the 49th percentile. If there is more than one data value closest to the 49th percentile, give any one answer.

    Answer: The value at or closest to the 49th percentile is: .
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the rank value from Question 1 to find the value in the data set closest to the 49th percentile.


    STEP: Use the rank of the 49th percentile to find the correct value in the data set
    [−1 point ⇒ 0 / 1 points left]

    Now we need to find the value in the data set which corresponds to the 49th percentile. To do this, we use the rank we calculated in Question 1. Remember that the rank tells us the position of a value in an ordered list. In this case we want to know which value has a rank of 6.39.

    In truth, the rank of a value must be an integer (because it represents a position). So we need to find the closest position value to 6.39.

    The rank value 6.39 is closest to sixth position in the data set. So the 49th percentile corresponds to 1.

    The value at or closest to the 49th percentile is 1.


    Submit your answer as:

ID is: 3186 Seed is: 9097

Applying percentile to a data set

The data set below has already been ordered for you. Find the item which corresponds to the 62nd percentile.

{28;11;9;7;5;6;14;15;18;19;21;21}
Answer: The item at or closest to the 62nd percentile is: .
one-of
type(numeric.noerror)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the rank of the item at the 62nd percentile.


STEP: Determine the rank of the item at the 62nd percentile
[−1 point ⇒ 1 / 2 points left]

We have a data set with 12 items in it, and the question asks us to determine the rank of the 62nd percentile in the data set. For this we use the formula:

r=p100(n1)+1

where:

  • p is the percentile value
  • n is the number of items in the data set
  • r is the rank (position) of the pth percentile in the data set

For this question, we have p=62 and n=12. Substitute these values into the formula above to find the rank.

r=p100(n1)+1=(62)100×((12)1)+1=62100×(11)+1=6.82+1=7.82

The rank of the 62nd percentile in the data set is 7.82.


STEP: Use the rank of the 62nd percentile to find the correct item in the data set
[−1 point ⇒ 0 / 2 points left]

Now we need to find the item in the data set which corresponds to the 62nd percentile. We use the rank to do this because the rank tells us the position of an item in an ordered list. In this case we want to know which item has a rank of 7.82.

But the rank of an item must be an integer because it represents a position. We need to find the closest position value to 7.82.

The rank value 7.82 is closest to eighth position in the data set. So the 62nd percentile corresponds to 15.

The item at the 62nd percentile is 15.


Submit your answer as:

ID is: 3186 Seed is: 5396

Applying percentile to a data set

The data set below has already been ordered for you. Which item is at the 0th percentile?

{c;h;h;h;j;j;k;l;m;p;q;r;u;x;z}
Answer: The item at or closest to the 0th percentile is: .
one-of
type(string)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You do not need to do any calculations to get this answer: just think about the meaning of the 0th percentile.


STEP: Determine the rank of the item at the 0th percentile
[−0 points ⇒ 1 / 1 points left]

For this question, it is not necessary to do any calculation in order to determine the item that is at the 0th percentile. The item at the 0th percentile is always the first one in an ordered data set. However we will carry out the working in order to confirm this fact.

We have a data set with 15 items in it, and the question asks us to determine the rank of the 0th percentile in the data set. For this we use the formula:

r=p100(n1)+1

where:

  • p is the percentile value
  • n is the number of items in the data set
  • r is the rank (position) of the pth percentile in the data set

For this question, we have p=0 and n=15. Substitute these values into the formula above to find the rank.

r=p100(n1)+1=(0)100×((15)1)+1=0100×(14)+1=0+1=1

The rank of the 0th percentile in the data set is 1.


STEP: Use the rank of the 0th percentile to find the correct item in the data set
[−1 point ⇒ 0 / 1 points left]

Now we need to find the item in the data set which corresponds to the 0th percentile. We use the rank to do this because the rank tells us the position of an item in an ordered list. In this case we want to know which item has a rank of 1.

In this case, we want the first letter in the data set.

The item at or closest to the 0th percentile is c.


Submit your answer as:

ID is: 3186 Seed is: 8637

Applying percentile to a data set

The data set below has already been ordered for you. Determine the item which corresponds to the 85th percentile.

{26;25;22;18;13;0;7;15;17;21;26}
Answer: The item at or closest to the 85th percentile is: .
one-of
type(numeric.noerror)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by finding the rank of the item at the 50th percentile.


STEP: Determine the rank of the item at the 85th percentile
[−1 point ⇒ 1 / 2 points left]

We have a data set with 11 items in it, and the question asks us to determine the rank of the 85th percentile in the data set. For this we use the formula:

r=p100(n1)+1

where:

  • p is the percentile value
  • n is the number of items in the data set
  • r is the rank (position) of the pth percentile in the data set

For this question, we have p=85 and n=11. Substitute these values into the formula above to find the rank.

r=p100(n1)+1=(85)100×((11)1)+1=85100×(10)+1=8.5+1=9.5

The rank of the 85th percentile in the data set is 9.5.


STEP: Use the rank of the 85th percentile to find the correct item in the data set
[−1 point ⇒ 0 / 2 points left]

Now we need to find the item in the data set which corresponds to the 85th percentile. We use the rank to do this because the rank tells us the position of an item in an ordered list. In this case we want to know which item has a rank of 9.5.

But the rank of an item must be an integer because it represents a position. We need to find the closest position value to 9.5.

The rank value 9.5 is exactly between the ninth and tenth positions. So the 85th percentile corresponds equally well to both 17 and other 21.

Either of these answers is acceptable for the item closest to the 85th percentile: 17 or 21.


Submit your answer as:

ID is: 2863 Seed is: 8787

The range

Consider the set of raw data:

{5;9;10;11;7;10;5;10;7;6;9}

What is the range of the data set?

Answer: Range:
numeric
2 attempts remaining
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The most useful step towards solving any data problem is to sort the data first.

{5;5;6;7;7;9;9;10;10;10;11}

The range of a set of data is the difference between the highest and lowest values in the set. In order to find the range, we should do the following:

  • Organise the data
  • Identify the lowest value
  • Identify the highest value and subtract them
Range=Highest value - lowest valueRange=115Range=6

Therefore, the range of the data set is 6.


Submit your answer as:

ID is: 2863 Seed is: 3399

The range

Consider the set of raw data:

{9;11;6;11;9;6;11;7;5}

What is the range of the data set?

Answer: Range:
numeric
2 attempts remaining
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The most useful step towards solving any data problem is to sort the data first.

{5;6;6;7;9;9;11;11;11}

The range of a set of data is the difference between the highest and lowest values in the set. In order to find the range, we should do the following:

  • Organise the data
  • Identify the lowest value
  • Identify the highest value and subtract them
Range=Highest value - lowest valueRange=115Range=6

Therefore, the range of the data set is 6.


Submit your answer as:

ID is: 2863 Seed is: 8165

The range

Consider the set of raw data:

{7;10;10;6;10;6;11}

What is the range of the data set?

Answer: Range:
numeric
2 attempts remaining
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The most useful step towards solving any data problem is to sort the data first.

{6;6;7;10;10;10;11}

The range of a set of data is the difference between the highest and lowest values in the set. In order to find the range, we should do the following:

  • Organise the data
  • Identify the lowest value
  • Identify the highest value and subtract them
Range=Highest value - lowest valueRange=116Range=5

Therefore, the range of the data set is 5.


Submit your answer as:

ID is: 855 Seed is: 8732

Calculating the range of a data set

A group of 20 students count the number of stones they each have. These are the data they collect:

11275156141111125416135181567

Calculate the range of values in the data set.

Answer: The range of the data is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by finding the maximum and minimum values in the data set. Then use them to find the range.


STEP: Find the maximum and minimum values and calculate the range
[−1 point ⇒ 0 / 1 points left]

The range of a set of data values is the difference between the largest and smallest values in the set. So we need to find the difference between the maximum and minimum data values.

Firstly, we find the maximum value in the data set.

Maximum value =18

Then we find the minimum value in the data set.

Minimum value =1

Finally, we calculate the range of the data set.

Range=MaximumMinimum=(18)(1)=17

So the range of the data is 17.


Submit your answer as:

ID is: 855 Seed is: 1037

Calculating the range of a data set

A group of 10 students count the number of coins they each have. These are the data they collect:

1433261236

Calculate the range of values in the data set.

Answer: The range of the data is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by finding the maximum and minimum values in the data set. Then use them to find the range.


STEP: Find the maximum and minimum values and calculate the range
[−1 point ⇒ 0 / 1 points left]

The range of a set of data values is the difference between the largest and smallest values in the set. So we need to find the difference between the maximum and minimum data values.

Firstly, we find the maximum value in the data set.

Maximum value =6

Then we find the minimum value in the data set.

Minimum value =1

Finally, we calculate the range of the data set.

Range=MaximumMinimum=(6)(1)=5

So the range of the data is 5.


Submit your answer as:

ID is: 855 Seed is: 1605

Calculating the range of a data set

A group of 10 students count the number of sweets they each have. These are the data they collect:

96108877434

Calculate the range of values in the data set.

Answer: The range of the data is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by finding the maximum and minimum values in the data set. Then use them to find the range.


STEP: Find the maximum and minimum values and calculate the range
[−1 point ⇒ 0 / 1 points left]

The range of a set of data values is the difference between the largest and smallest values in the set. So we need to find the difference between the maximum and minimum data values.

Firstly, we find the maximum value in the data set.

Maximum value =10

Then we find the minimum value in the data set.

Minimum value =3

Finally, we calculate the range of the data set.

Range=MaximumMinimum=(10)(3)=7

So the range of the data is 7.


Submit your answer as:

2. Variance & standard deviation


ID is: 3806 Seed is: 1738

Using x¯ and σ to compare data sets

The figure below shows two data sets:

Which of these statements correctly compares the mean and the standard deviation for these two data sets:

A Set 1 has a smaller mean but the standard deviation values are similar.
B The mean values are similar but Set 1 has a smaller standard deviation.
C The mean and standard deviation values for the sets are both similar.
D Neither the mean nor the standard deviation values for the two sets are similar.
Answer:

The most accurate description is statement .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetric data set. Use this as a guide to estimate the same information for each of the frequency polygons.


STEP: Compare the means and standard deviations of the graphs
[−1 point ⇒ 0 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetric data set:

The mean summarises a data set by pointing to the "balancing point" of the data. When the data are symmetric, the balancing point is in the middle. The standard deviation summarises the spread of the data around the mean. We can use the example above to roughly label the same information on the frequency polygons. We can do this because the shapes of the frequency polygons are similar to the example above.

What does this tell us about the values of the mean and the standard deviation?

  • The mean values for these sets are not similar. We do not know exactly what the mean values are equal to, but the graph shows that they are quite different. Set 1 has a larger mean than Set 2.
  • The standard deviation values are not similar. Set 1 is less spread out than Set 2.

The best statement is choice D.


Submit your answer as:

ID is: 3806 Seed is: 2911

Using x¯ and σ to compare data sets

The figure below shows two data sets:

Which of these statements correctly compares the mean and the standard deviation for these two data sets:

A Neither the mean nor the standard deviation values for the two sets are similar.
B Set 1 has a smaller mean but the standard deviation values are similar.
C The mean and standard deviation values for the sets are both similar.
D The mean values are similar but Set 1 has a smaller standard deviation.
Answer:

The most accurate description is statement .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetric data set. Use this as a guide to estimate the same information for each of the frequency polygons.


STEP: Compare the means and standard deviations of the graphs
[−1 point ⇒ 0 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetric data set:

The mean summarises a data set by pointing to the "balancing point" of the data. When the data are symmetric, the balancing point is in the middle. The standard deviation summarises the spread of the data around the mean. We can use the example above to roughly label the same information on the frequency polygons. We can do this because the shapes of the frequency polygons are similar to the example above.

What does this tell us about the values of the mean and the standard deviation?

  • The mean values for these sets are not similar. We do not know exactly what the mean values are equal to, but the graph shows that they are quite different. Set 1 has a larger mean than Set 2.
  • The standard deviation values are not similar. Set 1 is less spread out than Set 2.

The best statement is choice A.


Submit your answer as:

ID is: 3806 Seed is: 1459

Using x¯ and σ to compare data sets

The figure below shows two data sets:

Which of these statements correctly compares the mean and the standard deviation for these two data sets:

A The mean values are similar but Set 1 has a smaller standard deviation.
B The mean and standard deviation values for the sets are both similar.
C Set 1 has a larger mean but the standard deviation values are similar.
D Neither the mean nor the standard deviation values for the two sets are similar.
Answer:

The most accurate description is statement .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetric data set. Use this as a guide to estimate the same information for each of the frequency polygons.


STEP: Compare the means and standard deviations of the graphs
[−1 point ⇒ 0 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetric data set:

The mean summarises a data set by pointing to the "balancing point" of the data. When the data are symmetric, the balancing point is in the middle. The standard deviation summarises the spread of the data around the mean. We can use the example above to roughly label the same information on the frequency polygons. We can do this because the shapes of the frequency polygons are similar to the example above.

What does this tell us about the values of the mean and the standard deviation?

  • The mean values for these sets are similar. We do not know exactly what the mean values are equal to, but the graph shows that they are close to each other.
  • The standard deviation values are not similar. Set 1 is less spread out than Set 2.

The best statement is choice A.


Submit your answer as:

ID is: 4359 Seed is: 3501

Standard deviation from the mean

Adapted from DBE Nov 2015 Grade 11, P2, Q1
Maths formulas

The data shown below has a standard deviation of σ 49.52.

6419716956129601575399110
Answer:

The mean of the data is .

Is the value 53 more than one standard deviation below the mean?

one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You are given the standard deviation (σ). Find the mean (x¯) and the value of x¯σ to determine if 53 is less than this or not.


STEP: Find the mean
[−1 point ⇒ 2 / 3 points left]
The mean (x¯) is the sum of all the data values divided by the number of data values.
x¯=64+197+169+10=1,09410=109.4

STEP: Find the value of the expression
[−2 points ⇒ 0 / 3 points left]
The standard deviation (σ) is a measure of how much the values in a data set are spread out from the mean of the data. In other words, standard deviation tells us how tightly the numbers are clustered around the mean.
  • You are given the standard deviation in the question statement, σ 49.52.
  • You calculated the mean, x¯= 109.4.

Therefore:

x¯σ=109.449.52=59.88

53 is less than 59.88. Therefore the answer is yes, 53 is more than one standard deviation below the mean.


Submit your answer as: and

ID is: 4359 Seed is: 8263

Standard deviation from the mean

Adapted from DBE Nov 2015 Grade 11, P2, Q1
Maths formulas

The data shown below has a standard deviation of σ 64.1.

1992617816846149826116128
Answer:

The mean of the data is .

Is the value 26 within one standard deviation of the mean?

one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You are given the standard deviation (σ). Find the mean (x¯) and the values of x¯σ and x¯+σ to determine if 26 is within this range or not.


STEP: Find the mean
[−1 point ⇒ 2 / 3 points left]
The mean (x¯) is the sum of all the data values divided by the number of data values.
x¯=199+26+178+10=1,09810=109.8

STEP: Find the value of the expression
[−2 points ⇒ 0 / 3 points left]
The standard deviation (σ) is a measure of how much the values in a data set are spread out from the mean of the data. In other words, standard deviation tells us how tightly the numbers are clustered around the mean.
  • You are given the standard deviation in the question statement, σ 64.1.
  • You calculated the mean, x¯= 109.8.

Therefore:

x¯+σ=109.8+64.1=173.9

And:

x¯σ=109.864.1=45.7

We must decide if 26 lies between x¯σ and x¯+σ.

26 does not lie between 45.7 and 173.9. Therefore the answer is no, 26 is not within one standard deviation of the mean.


Submit your answer as: and

ID is: 4359 Seed is: 3181

Standard deviation from the mean

Adapted from DBE Nov 2015 Grade 11, P2, Q1
Maths formulas

The data shown below has a standard deviation of σ 54.15.

2813968167153921807647178
Answer:

The mean of the data is .

Is the value 180 more than one standard deviation above the mean?

one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You are given the standard deviation (σ). Find the mean (x¯) and the value of x¯+σ to determine if 180 is greater than this or not.


STEP: Find the mean
[−1 point ⇒ 2 / 3 points left]
The mean (x¯) is the sum of all the data values divided by the number of data values.
x¯=28+139+68+10=1,12810=112.8

STEP: Find the value of the expression
[−2 points ⇒ 0 / 3 points left]
The standard deviation (σ) is a measure of how much the values in a data set are spread out from the mean of the data. In other words, standard deviation tells us how tightly the numbers are clustered around the mean.
  • You are given the standard deviation in the question statement, σ 54.15.
  • You calculated the mean, x¯= 112.8.

Therefore:

x¯+σ=112.8+54.15=166.95

180 is more than 166.95. Therefore the answer is yes, 180 is more than one standard deviation above the mean.


Submit your answer as: and

ID is: 3785 Seed is: 3323

Sigma and the mean

The information below shows a set of 10 values. The mean of the values is x¯=4.5 and the standard deviation is σ2.2.

4;5;4;8;7;1;6;2;2;6

Give one value from the data set which is outside of one standard deviation of the mean.

INSTRUCTION: There might be more than one correct answer. You should give only one number.
Answer:

One number outside of one standard deviation of the mean is .

one-of
type(numeric.noerror)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by calculating 4.52.2 and 4.5+2.2.


STEP: Find one number outside of one standard deviation of the mean
[−1 point ⇒ 0 / 1 points left]

The standard deviation is a measure of how much the values in a data set are spread out from the mean of the data. In other words, standard deviation tells us how tightly the numbers are clustered around the mean.

We know that the data in this question have a mean of 4.5 and a standard deviation of 2.2. The picture below shows how these values combine to show the core of the data set:

The width of the zone shaded above comes directly from the standard deviation. Any data values inside that zone are "within one standard deviation of the mean." That means they are less than one standard deviation away from the mean. Values outside of the shaded zone are more than one standard deviation away from the mean. So to answer this question we need to determine the boundaries of the zone.

lower boundary=4.52.2=2.3upper boundary=4.5+2.2=6.7

Any number from the data set which is less than 2.3 or greater than 6.7 is more than one standard deviation away from the mean. So any of the numbers circled below are correct answers:

NOTE: Usually more than half, but not all, of the data values in a data set are within one standard deviation. The standard deviation measures how wide the core of the data set is. And there are usually some data values below and above this zone, like in the figure above.

These are the values outside of one standard deviation of the mean:

1,2,7,or 8

Submit your answer as:

ID is: 3785 Seed is: 1888

Sigma and the mean

The information below shows a set of 12 values. The mean of the values is x¯=5.5 and the standard deviation is σ3.

2;9;2;9;9;6;8;1;8;6;3;3

Give one value from the data set which is outside of one standard deviation of the mean.

INSTRUCTION: There might be more than one correct answer. You should give only one number.
Answer:

One number outside of one standard deviation of the mean is .

one-of
type(numeric.noerror)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by calculating 5.53 and 5.5+3.


STEP: Find one number outside of one standard deviation of the mean
[−1 point ⇒ 0 / 1 points left]

The standard deviation is a measure of how much the values in a data set are spread out from the mean of the data. In other words, standard deviation tells us how tightly the numbers are clustered around the mean.

We know that the data in this question have a mean of 5.5 and a standard deviation of 3. The picture below shows how these values combine to show the core of the data set:

The width of the zone shaded above comes directly from the standard deviation. Any data values inside that zone are "within one standard deviation of the mean." That means they are less than one standard deviation away from the mean. Values outside of the shaded zone are more than one standard deviation away from the mean. So to answer this question we need to determine the boundaries of the zone.

lower boundary=5.53=2.5upper boundary=5.5+3=8.5

Any number from the data set which is less than 2.5 or greater than 8.5 is more than one standard deviation away from the mean. So any of the numbers circled below are correct answers:

NOTE: Usually more than half, but not all, of the data values in a data set are within one standard deviation. The standard deviation measures how wide the core of the data set is. And there are usually some data values below and above this zone, like in the figure above.

These are the values outside of one standard deviation of the mean:

1,2,or 9

Submit your answer as:

ID is: 3785 Seed is: 2153

Sigma and the mean

The information below shows a set of 10 values. The mean of the values is x¯=6.3 and the standard deviation is σ2.1.

8;9;7;6;5;2;4;7;6;9

Give one value from the data set which is within one standard deviation of the mean.

INSTRUCTION: There might be more than one correct answer. You should give only one number.
Answer:

One number within one standard deviation of the mean is .

one-of
type(numeric.noerror)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by calculating 6.32.1 and 6.3+2.1.


STEP: Find one number within one standard deviation of the mean
[−1 point ⇒ 0 / 1 points left]

The standard deviation is a measure of how much the values in a data set are spread out from the mean of the data. In other words, standard deviation tells us how tightly the numbers are clustered around the mean.

We know that the data in this question have a mean of 6.3 and a standard deviation of 2.1. The picture below shows how these values combine to show the core of the data set:

The width of the zone shaded above comes directly from the standard deviation. Any data values inside that zone are "within one standard deviation of the mean." That means they are less than one standard deviation away from the mean. Values outside of the shaded zone are more than one standard deviation away from the mean. So to answer this question we need to determine the boundaries of the zone.

lower boundary=6.32.1=4.2upper boundary=6.3+2.1=8.4

Any number from the data set from 4.2 up to 8.4 is within one standard deviation of the mean. So any of the numbers circled below are correct answers:

NOTE: Usually more than half, but not all, of the data values in a data set are within one standard deviation. The standard deviation measures how wide the core of the data set is. And there are usually some data values below and above this zone, like in the figure above.

These are the values within one standard deviation of the mean:

5,6,7,or 8

Submit your answer as:

ID is: 3805 Seed is: 795

Matching the mean and standard deviation on a graph

The following histogram represents a set of data.

Which pair of statistics below best matches the data shown in the histogram?

A
x¯120σ25
B
x¯61σ50
C
x¯30σ50
D
x¯61σ8
Answer:

The best statistics for this graph are option .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetric data set. Use this as a guide to find the same information for each of the histograms.


STEP: Compare the values for the mean and standard deviation to the histogram
[−1 point ⇒ 0 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetrical data set:

The mean summarises a data set by pointing to the "balancing point" of the data. When the data are symmetric, the balancing point is in the middle. The standard deviation summarises the spread of the data around the mean. When the data are symmetric, about two-thirds of the data values will be within the interval (x¯σ;x¯+σ).

Since the histogram has a similar shape, we can use the example above to answer this question. Let's start by comparing the three choices given for the mean.

Only 61 appears to be a reasonable match. It sits near the middle of the histogram, which is what we expect for a fairly symmetrical data set. In contrast, the other possibilities are too far from the middle. Neither of them represent a reasonable balancing point for the data.

That leaves two possible correct choices. So we need to check how the σ values compare to the width of the histogram. The standard deviation should extend far enough from the mean to include most, but not all, of the data.

The value 8 is a reasonable size for the standard deviation. 50 is too wide: it extends beyond the central core of the histogram!

The mean and standard deviation values which agree best with the histogram are those in option D.


Submit your answer as:

ID is: 3805 Seed is: 6220

Matching the mean and standard deviation on a graph

The following histogram represents a set of data.

Which pair of statistics below best matches the data shown in the histogram?

A
x¯93σ37.5
B
x¯55σ37.5
C
x¯93σ13
D
x¯40σ75
Answer:

The best statistics for this graph are option .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetric data set. Use this as a guide to find the same information for each of the histograms.


STEP: Compare the values for the mean and standard deviation to the histogram
[−1 point ⇒ 0 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetrical data set:

The mean summarises a data set by pointing to the "balancing point" of the data. When the data are symmetric, the balancing point is in the middle. The standard deviation summarises the spread of the data around the mean. When the data are symmetric, about two-thirds of the data values will be within the interval (x¯σ;x¯+σ).

Since the histogram has a similar shape, we can use the example above to answer this question. Let's start by comparing the three choices given for the mean.

Only 93 appears to be a reasonable match. It sits near the middle of the histogram, which is what we expect for a fairly symmetrical data set. In contrast, the other possibilities are too far from the middle. Neither of them represent a reasonable balancing point for the data.

That leaves two possible correct choices. So we need to check how the σ values compare to the width of the histogram. The standard deviation should extend far enough from the mean to include most, but not all, of the data.

The value 13 is a reasonable size for the standard deviation. 37.5 is too wide: it extends beyond the central core of the histogram!

The mean and standard deviation values which agree best with the histogram are those in option C.


Submit your answer as:

ID is: 3805 Seed is: 151

Matching the mean and standard deviation on a graph

The following histogram represents a set of data.

Which pair of statistics below best matches the data shown in the histogram?

A
x¯175σ7.5
B
x¯112σ7.5
C
x¯112σ24
D
x¯40σ135
Answer:

The best statistics for this graph are option .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetric data set. Use this as a guide to find the same information for each of the histograms.


STEP: Compare the values for the mean and standard deviation to the histogram
[−1 point ⇒ 0 / 1 points left]

The graph below shows the mean and standard deviation labelled on a perfectly symmetrical data set:

The mean summarises a data set by pointing to the "balancing point" of the data. When the data are symmetric, the balancing point is in the middle. The standard deviation summarises the spread of the data around the mean. When the data are symmetric, about two-thirds of the data values will be within the interval (x¯σ;x¯+σ).

Since the histogram has a similar shape, we can use the example above to answer this question. Let's start by comparing the three choices given for the mean.

Only 112 appears to be a reasonable match. It sits near the middle of the histogram, which is what we expect for a fairly symmetrical data set. In contrast, the other possibilities are too far from the middle. Neither of them represent a reasonable balancing point for the data.

That leaves two possible correct choices. So we need to check how the σ values compare to the width of the histogram. The standard deviation should extend far enough from the mean to include most, but not all, of the data.

The value 24 is a reasonable size for the standard deviation. 7.5 is not wide enough: it is too small to include most of the data values.

The mean and standard deviation values which agree best with the histogram are those in option C.


Submit your answer as:

ID is: 3779 Seed is: 3041

Comparing how values are spread out

The figure below shows three different data sets graphed on number lines. Each dot represents a number in the data set. Answer the two questions below about these data sets.

  1. Which of the data sets is the most spread out?

    Answer:

    The most spread out set is .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Data values which are not spread out will be clumped up in tight groups. If the data are spread out, they will be further apart from each other.


    STEP: Identify the data which are most separated from each other
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the set with values which are the most spread out. This means we want the set with values stretched out as far as possible from each other, like the cricket team in this photo:

    Which of the number line diagrams is most similar to the picture above?

    • The least spread out is Set C. The dots for Set C are packed very close together with no gaps.
    • The most spread out set is Set A. The dots for Set A are stretched out far from each other with some gaps.
    • Set B is in between: it is more spread out than Set C but less spread out than Set A.

    The most spread out set is Set A.


    Submit your answer as:
  2. Which of the data sets has the largest standard deviation?

    TIP: Do not calculate the standard deviations. Your answer should be based on the spread of the data.
    Answer:

    The set with the largest standard deviation is .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation measures the spread of a data set. So think about the solution to Question 1.


    STEP: Identify the set with the largest spread
    [−1 point ⇒ 0 / 1 points left]

    Standard deviation measures the spread of a data set. If the numbers are bunched together like rugby players in a scrum, the standard deviation will be small. But if the numbers are spread far apart like the cricket team on the cricket pitch, the standard deviation will be larger.

    In Question 1 we found that Set C was the least spread out and Set A was the most spread out. Now we want to find the set with the largest standard deviation. So the answer must be Set A. In fact, the answer must be the same as the answer to Question 1 because the questions ask for the same thing!

    NOTE:

    The figure below shows the standard deviation values for each of the three sets in the question. You can see that the standard deviation is smallest for Set C because the data are packed very close to the mean with no open spaces. And it is largest for Set A because the data are stretched far from the mean with numerous gaps.

    The data set with the largest standard deviation is Set A.


    Submit your answer as:

ID is: 3779 Seed is: 2582

Comparing how values are spread out

The figure below shows three different data sets graphed on number lines. Each dot represents a number in the data set. Answer the two questions below about these data sets.

  1. Which of the data sets is the least spread out?

    Answer:

    The least spread out set is .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Data values which are not spread out will be clumped up in tight groups. If the data are spread out, they will be further apart from each other.


    STEP: Identify the data which are least separated from each other
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the set with values which are the least spread out. This means we want the set with values tightly packed next to each other, like the rugby players in this picture:

    Which of the number line diagrams is most similar to the picture above?

    • The least spread out is Set C. The dots for Set C are packed very close together with no gaps.
    • The most spread out set is Set A. The dots for Set A are stretched out far from each other with some gaps.
    • Set B is in between: it is more spread out than Set C but less spread out than Set A.

    The least spread out set is Set C.


    Submit your answer as:
  2. Which of the data sets has the smallest standard deviation?

    TIP: Do not calculate the standard deviations. Your answer should be based on the spread of the data.
    Answer:

    The set with the smallest standard deviation is .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation measures the spread of a data set. So think about the solution to Question 1.


    STEP: Identify the set with the smallest spread
    [−1 point ⇒ 0 / 1 points left]

    Standard deviation measures the spread of a data set. If the numbers are bunched together like rugby players in a scrum, the standard deviation will be small. But if the numbers are spread far apart like springbok in the veld, the standard deviation will be larger.

    In Question 1 we found that Set C was the least spread out and Set A was the most spread out. Now we want to find the set with the smallest standard deviation. So the answer must be Set C. In fact, the answer must be the same as the answer to Question 1 because the questions ask for the same thing!

    NOTE:

    The figure below shows the standard deviation values for each of the three sets in the question. You can see that the standard deviation is smallest for Set C because the data are packed very close to the mean with no open spaces. And it is largest for Set A because the data are stretched far from the mean with numerous gaps.

    The data set with the smallest standard deviation is Set C.


    Submit your answer as:

ID is: 3779 Seed is: 8118

Comparing how values are spread out

The figure below shows three different data sets graphed on number lines. Each dot represents a number in the data set. Answer the two questions below about these data sets.

  1. Which of the data sets is the least spread out?

    Answer:

    The least spread out set is .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Data values which are not spread out will be clumped up in tight groups. If the data are spread out, they will be further apart from each other.


    STEP: Identify the data which are least separated from each other
    [−1 point ⇒ 0 / 1 points left]

    We need to identify the set with values which are the least spread out. This means we want the set with values tightly packed next to each other, like the fish in this picture:

    Which of the number line diagrams is most similar to the picture above?

    • The least spread out is Set B. The dots for Set B are packed very close together with no gaps.
    • The most spread out set is Set A. The dots for Set A are spread out, with the smallest values pretty far from the largest values and with a couple of gaps.
    • Set C is in between: it is more spread out than Set B but less spread out than Set A.

    The least spread out set is Set B.


    Submit your answer as:
  2. Which of the data sets has the largest standard deviation?

    TIP: Do not calculate the standard deviations. Your answer should be based on the spread of the data.
    Answer:

    The set with the largest standard deviation is .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation measures the spread of a data set. So think about the solution to Question 1.


    STEP: Identify the set with the largest spread
    [−1 point ⇒ 0 / 1 points left]

    Standard deviation measures the spread of a data set. If the numbers are bunched together like fish packed in a can, the standard deviation will be small. But if the numbers are spread far apart like springbok in the veld, the standard deviation will be larger.

    In Question 1 we found that Set B was the least spread out and Set A was the most spread out. Now we want to find the set with the largest standard deviation. So the answer must be Set A. Interestingly, the answer cannot be the same as the answer to Question 1 because Question 1 asked for the opposite set - the one with the least spread out values.

    NOTE:

    The figure below shows the standard deviation values for each of the three sets in the question. You can see that the standard deviation is smallest for Set B because the data are packed very close to the mean with no open spaces. And it is largest for Set A because the data are spread out fairly far from the mean with a couple of gaps.

    The data set with the largest standard deviation is Set A.


    Submit your answer as:

ID is: 3907 Seed is: 7972

Mean and standard deviation

Adapted from DBE Nov 2016 Grade 12, P2, Q2.6
Maths formulas

The heights of 155 learners in a school are measured. The height of the shortest learner is 103 cm and the height of the tallest learner is 183 cm. The heights are represented in the histogram below.

The person taking the measurements only had a 150 cm measuring tape available. In order to compensate for the short measuring tape, he decided to mount the tape on a wall at a height of 100 cm above the ground. After recording the measurements he discovered that the tape was mounted at 94 cm above the ground instead of 100 cm.

  1. How does this error influence the mean of the data set? Select the best answer from the table of choices below.

    A The mean will be 6 cm too small.
    B The mean will be 94 cm too large.
    C The mean will be 94 cm too small.
    D The error has no effect on the mean.
    Answer:

    The correct answer is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    First think about how each number in the data set must change in order to correct the mistake in the measurement. What will those changes do to the data?


    STEP: Think about how adjusting the data values would impact the mean
    [−1 point ⇒ 0 / 1 points left]

    The mean of a data set is like a balancing point for the data: data values on one side of the mean are balanced out by data values on the other side of the mean. If the data values are all shifted by some amount, the mean must also shift. In fact, it will shift by the same amount.

    In this situation, the data values will all be 6 cm smaller than they should be. As a result, the mean will be 6 cm too small.

    The correct choice is option A.


    Submit your answer as:
  2. How does the error influence the standard deviation of the data set? Select the best answer from the table of choices below.

    A The standard deviation will be 6 cm too small.
    B The standard deviation will be 94 cm too large.
    C The error has no effect on the standard deviation.
    D The standard deviation will change by an unknown amount.
    Answer:

    The correct answer is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation summarises the spread of the data. How will the spread be affected by this mistake?


    STEP: Think about how adjusting the data values would impact the standard deviation
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation summarises the spread of a data set. If every one of the data values is shifted by the same amount, the spread will not change. Therefore, the measurement error has no impact on the standard deviation of the data.

    The correct choice is option C.


    Submit your answer as:

ID is: 3907 Seed is: 4795

Mean and standard deviation

Adapted from DBE Nov 2016 Grade 12, P2, Q2.6
Maths formulas

The heights of 160 learners in a school are measured. The height of the shortest learner is 1.12 m and the height of the tallest learner is 1.91 m. The heights are represented in the histogram below.

The person taking the measurements only had a 1.5 m measuring tape available. In order to compensate for the short measuring tape, he decided to mount the tape on a wall at a height of 1 m above the ground. After recording the measurements he discovered that the tape was mounted at 0.9 m above the ground instead of 1 m.

  1. How does this error influence the mean of the data set? Select the best answer from the table of choices below.

    A The mean will be 0.1 m too large.
    B The mean will be 0.1 m too small.
    C The mean will be 0.9 m too large.
    D The error has no effect on the mean.
    Answer:

    The correct answer is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    First think about how each number in the data set must change in order to correct the mistake in the measurement. What will those changes do to the data?


    STEP: Think about how adjusting the data values would impact the mean
    [−1 point ⇒ 0 / 1 points left]

    The mean of a data set is like a balancing point for the data: data values on one side of the mean are balanced out by data values on the other side of the mean. If the data values are all shifted by some amount, the mean must also shift. In fact, it will shift by the same amount.

    In this situation, the data values will all be 0.1 m smaller than they should be. As a result, the mean will be 0.1 m too small.

    The correct choice is option B.


    Submit your answer as:
  2. How does the error influence the standard deviation of the data set? Select the best answer from the table of choices below.

    A The standard deviation will be 0.1 m too small.
    B The standard deviation will be 0.9 m too large.
    C The error has no effect on the standard deviation.
    D The standard deviation will change by an unknown amount.
    Answer:

    The correct answer is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation summarises the spread of the data. How will the spread be affected by this mistake?


    STEP: Think about how adjusting the data values would impact the standard deviation
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation summarises the spread of a data set. If every one of the data values is shifted by the same amount, the spread will not change. Therefore, the measurement error has no impact on the standard deviation of the data.

    The correct choice is option C.


    Submit your answer as:

ID is: 3907 Seed is: 1818

Mean and standard deviation

Adapted from DBE Nov 2016 Grade 12, P2, Q2.6
Maths formulas

The heights of 175 learners in a school are measured. The height of the shortest learner is 94 cm and the height of the tallest learner is 181 cm. The heights are represented in the histogram below.

The person taking the measurements only had a 150 cm measuring tape available. In order to compensate for the short measuring tape, he decided to mount the tape on a wall at a height of 100 cm above the ground. After recording the measurements he discovered that the tape was mounted at 106 cm above the ground instead of 100 cm.

  1. How does this error influence the mean of the data set? Select the best answer from the table of choices below.

    A The mean will be 106 cm too small.
    B The mean will be 6 cm too large.
    C The mean will be 6 cm too small.
    D The error has no effect on the mean.
    Answer:

    The correct answer is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    First think about how each number in the data set must change in order to correct the mistake in the measurement. What will those changes do to the data?


    STEP: Think about how adjusting the data values would impact the mean
    [−1 point ⇒ 0 / 1 points left]

    The mean of a data set is like a balancing point for the data: data values on one side of the mean are balanced out by data values on the other side of the mean. If the data values are all shifted by some amount, the mean must also shift. In fact, it will shift by the same amount.

    In this situation, the data values will all be 6 cm larger than they should be. As a result, the mean will be 6 cm too large.

    The correct choice is option B.


    Submit your answer as:
  2. How does the error influence the standard deviation of the data set? Select the best answer from the table of choices below.

    A The standard deviation will be 106 cm too large.
    B The standard deviation will be 106 cm too small.
    C The error has no effect on the standard deviation.
    D The standard deviation will change by an unknown amount.
    Answer:

    The correct answer is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation summarises the spread of the data. How will the spread be affected by this mistake?


    STEP: Think about how adjusting the data values would impact the standard deviation
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation summarises the spread of a data set. If every one of the data values is shifted by the same amount, the spread will not change. Therefore, the measurement error has no impact on the standard deviation of the data.

    The correct choice is option C.


    Submit your answer as:

ID is: 3774 Seed is: 978

Standard deviation values

The following data set contains 10 values:

129;128;129;129;129;129;127;126;128;126

We can graph these values on a number line as follows:

One of the numbers below is the standard deviation of the data above (rounded to two decimal places). Which number must be the standard deviation?

A 128
B 3
C 0
D 1.18
TIP: Do not calculate the standard deviation. Instead, compare the numbers above to how much the data values are spread out on the number line.
Answer:

The standard deviation must be choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The standard deviation for a set of data summarises how the values are spread out around the mean of the data.


STEP: Identify which value is most reasonable for the spread of the data
[−1 point ⇒ 0 / 1 points left]

We need to select the correct value for the standard deviation of the numbers. Standard deviation tells us the spread of a data set. For example, if the values are clumped together like the flamingos in this photograph, the standard deviation must be small.

But if the values are spread apart like clouds in a mostly sunny sky, the standard deviation must be large.

The standard deviation shows us a zone around the mean which includes the core of the data set. For the data in this question the standard deviation looks like this:

Notice that the shaded zone includes most, but not all, of the data values. For example, 126 is outside of the shaded area. The standard deviation is the width of that shaded zone, and the answer must match the width of the arrows.

A ✘ 128
The spread of the data is much smaller than 128.
B ✘ 3
The answer should be somewhat smaller than 3 because the standard deviation does not span the entire data set.
C ✘ 0
The answer cannot be zero because the data are not all equal to each other.
D ✔ 1.18
This is the best match for the figure above.

Submit your answer as:

ID is: 3774 Seed is: 536

Standard deviation values

The following data set contains 10 values:

124;121;125;126;119;121;117;120;118;119

We can graph these values on a number line as follows:

One of the numbers below is the standard deviation of the data above (rounded to two decimal places). Which number must be the standard deviation?

A 2.9
B 121
C 9
D 126.8
TIP: Do not calculate the standard deviation. Instead, compare the numbers above to how much the data values are spread out on the number line.
Answer:

The standard deviation must be choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The standard deviation for a set of data summarises how the values are spread out around the mean of the data.


STEP: Identify which value is most reasonable for the spread of the data
[−1 point ⇒ 0 / 1 points left]

We need to select the correct value for the standard deviation of the numbers. Standard deviation tells us the spread of a data set. For example, if the values are clumped together like the fish in this picture, the standard deviation must be small.

But if the values are spread apart like springbok on the veld, the standard deviation must be large.

The standard deviation shows us a zone around the mean which includes the core of the data set. For the data in this question the standard deviation looks like this:

Notice that the shaded zone includes most, but not all, of the data values. For example, 126 is outside of the shaded area. The standard deviation is the width of that shaded zone, and the answer must match the width of the arrows.

A ✔ 2.9
This is the best match for the figure above.
B ✘ 121
The spread of the data is much smaller than 121.
C ✘ 9
The answer should be somewhat smaller than 9 because the standard deviation does not span the entire data set.
D ✘ 126.8
The spread of the data is much smaller than 126.8.

Submit your answer as:

ID is: 3774 Seed is: 9320

Standard deviation values

The following data set contains 9 values:

112;128;123;127;113;129;126;110;122

We can graph these values on a number line as follows:

One of the numbers below is the standard deviation of the data above (rounded to two decimal places). Which number must be the standard deviation?

A 135.17
B 1.34
C 7.03
D 19
TIP: Do not calculate the standard deviation. Instead, compare the numbers above to how much the data values are spread out on the number line.
Answer:

The standard deviation must be choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The standard deviation for a set of data summarises how the values are spread out around the mean of the data.


STEP: Identify which value is most reasonable for the spread of the data
[−1 point ⇒ 0 / 1 points left]

We need to select the correct value for the standard deviation of the numbers. Standard deviation tells us the spread of a data set. For example, if the values are clumped together like people at this bus stop, the standard deviation must be small.

But if the values are spread apart like clouds in a mostly sunny sky, the standard deviation must be large.

The standard deviation shows us a zone around the mean which includes the core of the data set. For the data in this question the standard deviation looks like this:

Notice that the shaded zone includes most, but not all, of the data values. For example, 112 is outside of the shaded area. The standard deviation is the width of that shaded zone, and the answer must match the width of the arrows.

A ✘ 135.17
The spread of the data is much smaller than 135.17.
B ✘ 1.34
This number is too small, because it implies the values are packed very close to the mean.
C ✔ 7.03
This is the best match for the figure above.
D ✘ 19
The answer should be somewhat smaller than 19 because the standard deviation does not span the entire data set.

Submit your answer as:

ID is: 3979 Seed is: 9446

Roll the dice

Adapted from DBE Nov 2015 Grade 12, P2, Q2
Maths formulas

A group of 34 learners each randomly rolled two dice once, and the sum of the values on the uppermost faces of the dice was recorded. The data are shown in the frequency table below.

Sum of the values Frequency
22
31
41
58
65
74
86
94
101
111
121
  1. Calculate the following for the data:

    INSTRUCTION: Round your answers to two decimal places.
    Answer:
    1. The mean of the data:
    2. The median of the data:
    3. The standard deviation of the data:
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    You may find it helpful to revise measures of central tendency or standard deviation in the Everything Maths textbook.


    STEP: Calculate the mean of the data
    [−2 points ⇒ 4 / 6 points left]

    To get the mean: multiply each data value with its frequency, add these products together, and then divide by the total number of students.

    x¯=2(2)+3(1)++12(1)34=22634=6.64705...6.65

    STEP: Calculate the median of the data
    [−2 points ⇒ 2 / 6 points left]

    The median value of a data set is the value in the central position, when the data have been arranged from the lowest to the highest value. Since our data set has 34 data values (an even number), the median is the average of the 17th and 18th data points.

    median=6+72=6.5

    STEP: Calculate the standard deviation of the data
    [−2 points ⇒ 0 / 6 points left]
    TIP: You should use your calculator to determine the standard deviation.
    σ=2.29939...2.3

    Submit your answer as: andand
  2. Determine the number of times that the sum of the recorded values of the dice is within one standard deviation from the mean.

    Answer:

    The number of data points within one standard deviation of the mean is .

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The upper and lower bounds of the interval will be x¯+σ and x¯σ respectively.


    STEP: Calculate the upper and lower bounds of the interval
    [−2 points ⇒ 1 / 3 points left]

    The bounds of the interval within one standard deviation of the mean will be

    upper bound=x¯+σ=6.64705...+2.29939...=8.94644...
    lower bound=x¯σ=6.64705...2.29939...=4.34766...

    STEP: Determine the number of data points in the interval
    [−1 point ⇒ 0 / 3 points left]

    Now we need to count the number of data points that lie within this interval. The values we got for the lower and upper bounds are not integers, but since the data we are working with are discrete, we want all of the values between 5 and 8.

    number of points=f5+f6+f7+f8=8+5+4+6=23

    Submit your answer as:

ID is: 3979 Seed is: 3328

Roll the dice

Adapted from DBE Nov 2015 Grade 12, P2, Q2
Maths formulas

A group of 33 learners each randomly rolled two dice once, and the sum of the values on the uppermost faces of the dice was recorded. The data are shown in the frequency table below.

Sum of the values Frequency
21
30
41
52
67
77
87
91
103
113
121
  1. Calculate the following for the data:

    INSTRUCTION: Round your answers to two decimal places.
    Answer:
    1. The mean of the data:
    2. The median of the data:
    3. The standard deviation of the data:
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    You may find it helpful to revise measures of central tendency or standard deviation in the Everything Maths textbook.


    STEP: Calculate the mean of the data
    [−2 points ⇒ 4 / 6 points left]

    To get the mean: multiply each data value with its frequency, add these products together, and then divide by the total number of students.

    x¯=2(1)+3(0)++12(1)33=24733=7.48484...7.48

    STEP: Calculate the median of the data
    [−2 points ⇒ 2 / 6 points left]

    The median value of a data set is the value in the central position, when the data have been arranged from the lowest to the highest value. Since our data set has 33 data values (an odd number), the median is the 17th data point.

    median=7

    STEP: Calculate the standard deviation of the data
    [−2 points ⇒ 0 / 6 points left]
    TIP: You should use your calculator to determine the standard deviation.
    σ=2.14788...2.15

    Submit your answer as: andand
  2. Determine the number of times that the sum of the recorded values of the dice is within one standard deviation from the mean.

    Answer:

    The number of data points within one standard deviation of the mean is .

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The upper and lower bounds of the interval will be x¯+σ and x¯σ respectively.


    STEP: Calculate the upper and lower bounds of the interval
    [−2 points ⇒ 1 / 3 points left]

    The bounds of the interval within one standard deviation of the mean will be

    upper bound=x¯+σ=7.48484...+2.14788...=9.63273...
    lower bound=x¯σ=7.48484...2.14788...=5.33696...

    STEP: Determine the number of data points in the interval
    [−1 point ⇒ 0 / 3 points left]

    Now we need to count the number of data points that lie within this interval. The values we got for the lower and upper bounds are not integers, but since the data we are working with are discrete, we want all of the values between 6 and 9.

    number of points=f6+f7+f8+f9=7+7+7+1=22

    Submit your answer as:

ID is: 3979 Seed is: 664

Roll the dice

Adapted from DBE Nov 2015 Grade 12, P2, Q2
Maths formulas

A group of 33 learners each randomly rolled two dice once, and the sum of the values on the uppermost faces of the dice was recorded. The data are shown in the frequency table below.

Sum of the values Frequency
22
30
44
51
65
78
85
93
102
112
121
  1. Calculate the following for the data:

    INSTRUCTION: Round your answers to two decimal places.
    Answer:
    1. The mean of the data:
    2. The median of the data:
    3. The standard deviation of the data:
    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    You may find it helpful to revise measures of central tendency or standard deviation in the Everything Maths textbook.


    STEP: Calculate the mean of the data
    [−2 points ⇒ 4 / 6 points left]

    To get the mean: multiply each data value with its frequency, add these products together, and then divide by the total number of students.

    x¯=2(2)+3(0)++12(1)33=23233=7.03030...7.03

    STEP: Calculate the median of the data
    [−2 points ⇒ 2 / 6 points left]

    The median value of a data set is the value in the central position, when the data have been arranged from the lowest to the highest value. Since our data set has 33 data values (an odd number), the median is the 17th data point.

    median=7

    STEP: Calculate the standard deviation of the data
    [−2 points ⇒ 0 / 6 points left]
    TIP: You should use your calculator to determine the standard deviation.
    σ=2.38028...2.38

    Submit your answer as: andand
  2. Determine the number of times that the sum of the recorded values of the dice is within one standard deviation from the mean.

    Answer:

    The number of data points within one standard deviation of the mean is .

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The upper and lower bounds of the interval will be x¯+σ and x¯σ respectively.


    STEP: Calculate the upper and lower bounds of the interval
    [−2 points ⇒ 1 / 3 points left]

    The bounds of the interval within one standard deviation of the mean will be

    upper bound=x¯+σ=7.03030...+2.38028...=9.41058...
    lower bound=x¯σ=7.03030...2.38028...=4.65001...

    STEP: Determine the number of data points in the interval
    [−1 point ⇒ 0 / 3 points left]

    Now we need to count the number of data points that lie within this interval. The values we got for the lower and upper bounds are not integers, but since the data we are working with are discrete, we want all of the values between 5 and 9.

    number of points=f5+f6+f7+f8+f9=1+5+8+5+3=22

    Submit your answer as:

ID is: 3810 Seed is: 197

How data values impact σ

Answer the three questions below about this set of numbers:

1;3;3;5;5;9x¯=4,3˙σ=2.49443...
  1. Which number in the set has the largest impact on the value of σ?

    INSTRUCTION: If there is more than one correct answer, you should give any one of them.
    Answer:

    The number with the largest impact on σ is .

    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation is about the spread of numbers from the mean. So numbers furthest from the mean have the biggest impact on σ and numbers closest to the mean have the smallest impact.


    STEP: Compare the distances of the numbers from the mean value
    [−1 point ⇒ 0 / 1 points left]

    Standard deviation measures how much a set of values is spread out from the mean of those values. In other words, it summarises the distances of the values from the mean. The graph below shows the distances from the mean of the data to each of the six data values.

    Numbers closest to the mean make the smallest contribution to the standard deviation. And numbers furthest away from the mean make the largest contribution. We want the numbers which have the largest impact on σ, so we need to find the numbers which are furthest from the mean.

    Based on the diagram we can see that 9 is furthest from the mean.

    NOTE:

    Why do larger distances from the mean affect the mean more than shorter distances? The answer comes directly from the formula for the standard deviation. Suppose a set contains three numbers a, b, and c. Then the formula would include the following sum:

    (ax¯)2+(bx¯)2+(cx¯)2

    For these three terms, the largest one will come from whichever data value is furthest from x¯. So it adds the most to the sum and hence has the biggest impact on the standard deviation. Similarly, the smallest term will come from whichever data value is closest to x¯. So it adds the least to the sum, having the smallest impact.

    The number with the largest impact on σ is 9.


    Submit your answer as:
  2. The formula for the standard deviation is:

    σ=(xx¯)2N

    In this formula, why is the quantity (xx¯) squared?

    (xx¯) is squared because...
    A it prevents division by zero (prevents undefined answers).
    B it makes sure that we must use FOIL to square the binomial.
    C it changes negative distances into positive values.
    D mathematicians know students love complicated calculations.
    Answer: The correct option is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The question asks why the formula for the standard deviation includes (xx¯)2 rather than (xx¯). The answer is related to this statement: n20.


    STEP: Identify the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks why the formula for the standard deviation includes (xx¯)2 rather than (xx¯). The square is crucial in this formula because it makes sure that every distance represented by (xx¯) ends up being positive.

    Remember that the expression (xx¯) represents how far each value is from the mean. For every value greater than the mean, (xx¯) will be positive. But for every value less than the mean, (xx¯) will be negative. Without the square, those positive and negative values would cancel each other out in the σ calculation. That cancellation would reduce the size of σ, giving us the incorrect impression that the values were not so spread out.

    The square avoids that cancellation by changing all negative values into positive values. So numbers above the mean and below the mean all add to (increase) the standard deviation value. In other words, the square ensures that when we calculate σ, numbers above and below the mean are treated equally.

    The standard deviation formula includes the square because it changes negative distances into positive values. So the correct choice is C.


    Submit your answer as:
  3. Suppose 3 in the original data set changes to 2. This will change the mean to 3,5 as you can see here:

    1;3;3;5;5;9x¯=4,3˙σ=2.49443...Change 3 to 22;1;3;5;5;9x¯new=3,5

    Will this change cause the standard deviation to increase, decrease, or stay the same?

    TIP: Do not calculate the new standard deviation. Use what you know about the concept of standard deviation.
    Answer: The standard deviation will .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The answer depends on the change in distance of the number from the mean of the data.


    STEP: Compare the distances of the numbers from the mean values
    [−1 point ⇒ 0 / 1 points left]

    We need to determine how the standard deviation of the set will change if the value 3 changes to 2. As with Question 1, this comes down to how the distances of the data values from the mean change.

    In the figure below, x¯ is the mean of the original data set and x¯new is the mean after the change. We need to compare the distances of the original and final values to these means to see if the distance increased or decreased.

    The picture shows that the distance from x¯new to 2 is bigger than the distance from x¯ to 3. So the size of the (xx¯) term in the σ calculation will get bigger. And that will cause the σ value to increase.

    NOTE: We ignored a detail in the explanation above. When the mean changes, the distances from the mean changes for all six values. And all those changes impact the standard deviation. However, for the five numbers in the set which stayed the same, those changes are quite small and tend to balance each other out. The change in distance when 3 changes to 2 is a much larger change, and it has the strongest impact on the change in σ.

    If we change the number 3 to 2, the standard deviation will increase.


    Submit your answer as:

ID is: 3810 Seed is: 4491

How data values impact σ

Answer the three questions below about this set of numbers:

2;2;7;9;9;9x¯=6,3˙σ=3.14466...
  1. Which number in the set has the smallest impact on the value of σ?

    INSTRUCTION: If there is more than one correct answer, you should give any one of them.
    Answer:

    The number with the smallest impact on σ is .

    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation is about the spread of numbers from the mean. So numbers furthest from the mean have the biggest impact on σ and numbers closest to the mean have the smallest impact.


    STEP: Compare the distances of the numbers from the mean value
    [−1 point ⇒ 0 / 1 points left]

    Standard deviation measures how much a set of values is spread out from the mean of those values. In other words, it summarises the distances of the values from the mean. The graph below shows the distances from the mean of the data to each of the six data values.

    Numbers closest to the mean make the smallest contribution to the standard deviation. And numbers furthest away from the mean make the largest contribution. We want the numbers which have the smallest impact on σ, so we need to find the numbers which are closest to the mean.

    Based on the diagram we can see that 7 is closest to the mean.

    NOTE:

    Why do larger distances from the mean affect the mean more than shorter distances? The answer comes directly from the formula for the standard deviation. Suppose a set contains three numbers a, b, and c. Then the formula would include the following sum:

    (ax¯)2+(bx¯)2+(cx¯)2

    For these three terms, the largest one will come from whichever data value is furthest from x¯. So it adds the most to the sum and hence has the biggest impact on the standard deviation. Similarly, the smallest term will come from whichever data value is closest to x¯. So it adds the least to the sum, having the smallest impact.

    The number with the smallest impact on σ is 7.


    Submit your answer as:
  2. The formula for the standard deviation is:

    σ=(xx¯)2N

    In this formula, why is the quantity (xx¯) squared?

    (xx¯) is squared because...
    A it cancels out the subtraction effect.
    B it makes sure that we must use FOIL to square the binomial.
    C it changes negative distances into positive values.
    D exponents of 2 make everything better.
    Answer: The correct option is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The question asks why the formula for the standard deviation includes (xx¯)2 rather than (xx¯). The answer is related to this statement: n20.


    STEP: Identify the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks why the formula for the standard deviation includes (xx¯)2 rather than (xx¯). The square is crucial in this formula because it makes sure that every distance represented by (xx¯) ends up being positive.

    Remember that the expression (xx¯) represents how far each value is from the mean. For every value greater than the mean, (xx¯) will be positive. But for every value less than the mean, (xx¯) will be negative. Without the square, those positive and negative values would cancel each other out in the σ calculation. That cancellation would reduce the size of σ, giving us the incorrect impression that the values were not so spread out.

    The square avoids that cancellation by changing all negative values into positive values. So numbers above the mean and below the mean all add to (increase) the standard deviation value. In other words, the square ensures that when we calculate σ, numbers above and below the mean are treated equally.

    The standard deviation formula includes the square because it changes negative distances into positive values. So the correct choice is C.


    Submit your answer as:
  3. Suppose 9 in the original data set changes to 7. This will change the mean to 6 as you can see here:

    2;2;7;9;9;9x¯=6,3˙σ=3.14466...Change 9 to 72;2;7;7;9;9x¯new=6

    Will this change cause the standard deviation to increase, decrease, or stay the same?

    TIP: Do not calculate the new standard deviation. Use what you know about the concept of standard deviation.
    Answer: The standard deviation will .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The answer depends on the change in distance of the number from the mean of the data.


    STEP: Compare the distances of the numbers from the mean values
    [−1 point ⇒ 0 / 1 points left]

    We need to determine how the standard deviation of the set will change if the value 9 changes to 7. As with Question 1, this comes down to how the distances of the data values from the mean change.

    In the figure below, x¯ is the mean of the original data set and x¯new is the mean after the change. We need to compare the distances of the original and final values to these means to see if the distance increased or decreased.

    The picture shows that the distance from x¯new to 7 is smaller than the distance from x¯ to 9. So the size of the (xx¯) term in the σ calculation will get smaller. And that will cause the σ value to decrease.

    NOTE: We ignored a detail in the explanation above. When the mean changes, the distances from the mean changes for all six values. And all those changes impact the standard deviation. However, for the five numbers in the set which stayed the same, those changes are quite small and tend to balance each other out. The change in distance when 9 changes to 7 is a much larger change, and it has the strongest impact on the change in σ.

    If we change the number 9 to 7, the standard deviation will decrease.


    Submit your answer as:

ID is: 3810 Seed is: 4990

How data values impact σ

Answer the three questions below about this set of numbers:

2;3;4;5;5;7x¯=4,3˙σ=1.59861...
  1. Which number in the set has the smallest impact on the value of σ?

    INSTRUCTION: If there is more than one correct answer, you should give any one of them.
    Answer:

    The number with the smallest impact on σ is .

    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation is about the spread of numbers from the mean. So numbers furthest from the mean have the biggest impact on σ and numbers closest to the mean have the smallest impact.


    STEP: Compare the distances of the numbers from the mean value
    [−1 point ⇒ 0 / 1 points left]

    Standard deviation measures how much a set of values is spread out from the mean of those values. In other words, it summarises the distances of the values from the mean. The graph below shows the distances from the mean of the data to each of the six data values.

    Numbers closest to the mean make the smallest contribution to the standard deviation. And numbers furthest away from the mean make the largest contribution. We want the numbers which have the smallest impact on σ, so we need to find the numbers which are closest to the mean.

    Based on the diagram we can see that 4 is closest to the mean.

    NOTE:

    Why do larger distances from the mean affect the mean more than shorter distances? The answer comes directly from the formula for the standard deviation. Suppose a set contains three numbers a, b, and c. Then the formula would include the following sum:

    (ax¯)2+(bx¯)2+(cx¯)2

    For these three terms, the largest one will come from whichever data value is furthest from x¯. So it adds the most to the sum and hence has the biggest impact on the standard deviation. Similarly, the smallest term will come from whichever data value is closest to x¯. So it adds the least to the sum, having the smallest impact.

    The number with the smallest impact on σ is 4.


    Submit your answer as:
  2. The formula for the standard deviation is:

    σ=(xx¯)2N

    In this formula, why is the quantity (xx¯) squared?

    (xx¯) is squared because...
    A it changes negative distances into positive values.
    B the formula must be quadratic to make a parabola.
    C it cancels out the subtraction effect.
    D exponents of 2 make everything better.
    Answer: The correct option is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The question asks why the formula for the standard deviation includes (xx¯)2 rather than (xx¯). The answer is related to this statement: n20.


    STEP: Identify the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks why the formula for the standard deviation includes (xx¯)2 rather than (xx¯). The square is crucial in this formula because it makes sure that every distance represented by (xx¯) ends up being positive.

    Remember that the expression (xx¯) represents how far each value is from the mean. For every value greater than the mean, (xx¯) will be positive. But for every value less than the mean, (xx¯) will be negative. Without the square, those positive and negative values would cancel each other out in the σ calculation. That cancellation would reduce the size of σ, giving us the incorrect impression that the values were not so spread out.

    The square avoids that cancellation by changing all negative values into positive values. So numbers above the mean and below the mean all add to (increase) the standard deviation value. In other words, the square ensures that when we calculate σ, numbers above and below the mean are treated equally.

    The standard deviation formula includes the square because it changes negative distances into positive values. So the correct choice is A.


    Submit your answer as:
  3. Suppose 4 in the original data set changes to 10. This will change the mean to 5,3˙ as you can see here:

    2;3;4;5;5;7x¯=4,3˙σ=1.59861...Change 4 to 102;3;5;5;7;10x¯new=5,3˙

    Will this change cause the standard deviation to increase, decrease, or stay the same?

    TIP: Do not calculate the new standard deviation. Use what you know about the concept of standard deviation.
    Answer: The standard deviation will .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The answer depends on the change in distance of the number from the mean of the data.


    STEP: Compare the distances of the numbers from the mean values
    [−1 point ⇒ 0 / 1 points left]

    We need to determine how the standard deviation of the set will change if the value 4 changes to 10. As with Question 1, this comes down to how the distances of the data values from the mean change.

    In the figure below, x¯ is the mean of the original data set and x¯new is the mean after the change. We need to compare the distances of the original and final values to these means to see if the distance increased or decreased.

    The picture shows that the distance from x¯new to 10 is bigger than the distance from x¯ to 4. So the size of the (xx¯) term in the σ calculation will get bigger. And that will cause the σ value to increase. (This distance changes sign because 4 is on the left side of the mean, while 10 is on the right side. But as we saw in Question 2, the square in the standard deviation formula makes sure that positive and negative values both make a positive contribution to the σ value.)

    NOTE: We ignored a detail in the explanation above. When the mean changes, the distances from the mean changes for all six values. And all those changes impact the standard deviation. However, for the five numbers in the set which stayed the same, those changes are quite small and tend to balance each other out. The change in distance when 4 changes to 10 is a much larger change, and it has the strongest impact on the change in σ.

    If we change the number 4 to 10, the standard deviation will increase.


    Submit your answer as:

ID is: 3809 Seed is: 6758

How do the data values impact σ?

  1. Two data sets are shown below. The second set has one number missing which is represented by n. The sets have the same standard deviation, σ=2.94859....

    Set 1:_1;5;4;9;7;2;6;0;6;2;2σ1=2.94859...
    Set 2:_8;6;n;13;12;5;12;15;8;11;8σ2=2.94859...

    Determine one possible value for the missing number, n.

    Answer: n=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Draw a number line graph for each set. Then compare them to each other.


    STEP: Compare the sets using number lines graphs
    [−1 point ⇒ 1 / 2 points left]

    Standard deviation summarises the spread of a data set around the mean. If two sets have equal σ values, they must have equivalent spreads. So we need to compare the structures of the sets. How can we make that comparison?

    Let's draw number line graphs for the sets. The second graph will have one less dot than the first one because we do not know where to put n yet.

    The structure of the dots is almost identical. This is a reflection of the fact that the sets have equal σ-values. It also emphasises that the shift between sets does not impact the standard deviation.


    STEP: Determine the position and value of n
    [−1 point ⇒ 0 / 2 points left]

    We can locate the position of n because we know that the structure of the sets must match. Comparing the number line diagrams above, we can see that Set 2 is missing a value corresponding to 4 from Set 1. 3

    We can read the value of n from the number line now.

    The value of n is 10.


    Submit your answer as:
  2. The data below have a mean of 69.38 and a standard deviation of 2.29.

    66;67;67;70;70;71;71;73x¯=69.38σ=2.29

    If 70 is removed, would the value of σ increase, decrease, or stay the same?

    TIP: Do not calculate the new standard deviation. Use what you know about the concept of standard deviation.
    Answer: The value of σ would .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Is 70 within one σ of the mean value? Or is it outside of that boundary?


    STEP: Think about the distance of 70 from the mean
    [−1 point ⇒ 0 / 1 points left]

    To answer this question, we need to compare 70 to the boundary defined by σ around the mean. If we remove a number which is inside this boundary, σ should increase. That happens because we are removing a number with below-average spread and the spread will increase. But if we remove a number which is outside this boundary, σ should decrease. That happens because we are taking away a number which is far from the mean which reduces the overall spread.

    It is easiest to see this on a number line graph of the data values.

    The graph shows that the removed data value 70 is well within the one σ boundary around x¯. So removing it from the data set will increase the overall spread of the data set.

    If 70 is removed the value of σ will increase.


    Submit your answer as:
  3. The following data set has ten values in total with a single unknown value, x:

    12;15;8;13;14;13;5;15;7;x

    The graph below shows the functions σ(x) and n¯(x) over a domain starting at x=0 and reaching an unspecified maximum value, xmax. The value x=p is marked on the horizontal axis.

    To approximately one decimal place, what is the value of p?

    Answer: p
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Do not try to calculate the answer. You can read the answer from the graph.


    STEP: Apply the concept of standard deviation from the mean
    [−1 point ⇒ 0 / 1 points left]

    We need to find a way to determine the value of p from the graph. Do you notice anything special about x=p?

    The only noteworthy thing about x=p is that it is where the standard deviation function has a minimum value. It turns out that we can use this fact to determine p!

    As x increases from 0, σ(x) decreases. This must be because the value of x is getting closer to the mean of the data set: that decreases the overall spread of the data set, pushing the value of σ down. But then, after x=p, σ(x) turns around and starts to increase. This means x is now getting further from the mean of the data, not closer, which increases the overall spread of the data. So at x=p (at the minimum value of σ(x)), x must be as close as possible to the mean.

    What is the closest x can get to the mean? It can be equal to the mean! So at x=p, x must be equal to the mean. This is shown on the graph below.

    The minimum value of σ(x) occurs when x is equal to the mean. The mean is labelled on the right axis, so that is where we need to read it off: p11.3.

    NOTE:

    It is possible to calculate the exact answer. We know that p is the value for which x is equal to the mean.

    x=x+(12+15+8+13+14+13+5+15+7)1010x=x+1029x=102x=102911.3

    This answer is consistent with the estimation we found from the graph above.

    Check out the meaning of the fraction, 1029: it is the sum of the nine known data values (the ten values without x) divided by 9. That is the mean of those nine values. This is really cool: it tells us that to get the smallest standard deviation possible for all ten numbers, x should be equal to the mean of the other nine numbers. That is deep: if x is equal to the mean of those nine numbers, then x itself adds nothing more to the standard deviation of those numbers. Maths is amazing - that is worth at least two exclamation marks!!

    The value of p is approximately 11.3.


    Submit your answer as:

ID is: 3809 Seed is: 1021

How do the data values impact σ?

  1. Two data sets are shown below. The second set has one number missing which is represented by a. The sets have the same standard deviation, σ=2.00997....

    Set 1:_3;2;3;5;0;4;3;0;1;5σ1=2.00997...
    Set 2:_3;0;2;4;1;a;3;0;0;2σ2=2.00997...

    Determine one possible value for the missing number, a.

    Answer: a=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Draw a number line graph for each set. Then compare them to each other.


    STEP: Compare the sets using number lines graphs
    [−1 point ⇒ 1 / 2 points left]

    Standard deviation summarises the spread of a data set around the mean. If two sets have equal σ values, they must have equivalent spreads. So we need to compare the structures of the sets. How can we make that comparison?

    Let's draw number line graphs for the sets. The second graph will have one less dot than the first one because we do not know where to put a yet.

    The structure of the dots is almost identical. This is a reflection of the fact that the sets have equal σ-values. It also emphasises that the shift between sets does not impact the standard deviation.


    STEP: Determine the position and value of a
    [−1 point ⇒ 0 / 2 points left]

    We can locate the position of a because we know that the structure of the sets must match. Comparing the number line diagrams above, we can see that Set 2 is missing a value corresponding to 4 from Set 1. 2

    We can read the value of a from the number line now.

    The value of a is 1.


    Submit your answer as:
  2. The data below have a mean of 91.38 and a standard deviation of 2.83.

    86;88;90;93;93;93;94;94x¯=91.38σ=2.83

    If 86 is removed, would the value of σ increase, decrease, or stay the same?

    TIP: Do not calculate the new standard deviation. Use what you know about the concept of standard deviation.
    Answer: The value of σ would .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Is 86 within one σ of the mean value? Or is it outside of that boundary?


    STEP: Think about the distance of 86 from the mean
    [−1 point ⇒ 0 / 1 points left]

    To answer this question, we need to compare 86 to the boundary defined by σ around the mean. If we remove a number which is inside this boundary, σ should increase. That happens because we are removing a number with below-average spread and the spread will increase. But if we remove a number which is outside this boundary, σ should decrease. That happens because we are taking away a number which is far from the mean which reduces the overall spread.

    It is easiest to see this on a number line graph of the data values.

    The graph shows that the removed data value 86 is well outside the one σ boundary around x¯. So removing it from the data set will decrease the overall spread of the data set.

    If 86 is removed the value of σ will decrease.


    Submit your answer as:
  3. The following data set has ten values in total with a single unknown value, x:

    20;23;19;16;16;20;18;23;20;x

    The graph below shows the functions σ(x) and n¯(x) over a domain starting at x=0 and reaching an unspecified maximum value, xmax. The value x=p is marked on the horizontal axis.

    To approximately one decimal place, what is the value of p?

    Answer: p
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Do not try to calculate the answer. You can read the answer from the graph.


    STEP: Apply the concept of standard deviation from the mean
    [−1 point ⇒ 0 / 1 points left]

    We need to find a way to determine the value of p from the graph. Do you notice anything special about x=p?

    The only noteworthy thing about x=p is that it is where the standard deviation function has a minimum value. It turns out that we can use this fact to determine p!

    As x increases from 0, σ(x) decreases. This must be because the value of x is getting closer to the mean of the data set: that decreases the overall spread of the data set, pushing the value of σ down. But then, after x=p, σ(x) turns around and starts to increase. This means x is now getting further from the mean of the data, not closer, which increases the overall spread of the data. So at x=p (at the minimum value of σ(x)), x must be as close as possible to the mean.

    What is the closest x can get to the mean? It can be equal to the mean! So at x=p, x must be equal to the mean. This is shown on the graph below.

    The minimum value of σ(x) occurs when x is equal to the mean. The mean is labelled on the right axis, so that is where we need to read it off: p19.4.

    NOTE:

    It is possible to calculate the exact answer. We know that p is the value for which x is equal to the mean.

    x=x+(20+23+19+16+16+20+18+23+20)1010x=x+1759x=175x=175919.4

    This answer is consistent with the estimation we found from the graph above.

    Check out the meaning of the fraction, 1759: it is the sum of the nine known data values (the ten values without x) divided by 9. That is the mean of those nine values. This is really cool: it tells us that to get the smallest standard deviation possible for all ten numbers, x should be equal to the mean of the other nine numbers. That is deep: if x is equal to the mean of those nine numbers, then x itself adds nothing more to the standard deviation of those numbers. Maths is amazing - that is worth at least two exclamation marks!!

    The value of p is approximately 19.4.


    Submit your answer as:

ID is: 3809 Seed is: 5236

How do the data values impact σ?

  1. Two data sets are shown below. The second set has one number missing which is represented by a. The sets have the same standard deviation, σ=2.19215....

    Set 1:_8;10;11;14;14;11;14;14;15;9;13;13σ1=2.19215...
    Set 2:_20;17;20;15;21;20;17;16;19;20;a;19σ2=2.19215...

    Determine one possible value for the missing number, a.

    Answer: a=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Draw a number line graph for each set. Then compare them to each other.


    STEP: Compare the sets using number lines graphs
    [−1 point ⇒ 1 / 2 points left]

    Standard deviation summarises the spread of a data set around the mean. If two sets have equal σ values, they must have equivalent spreads. So we need to compare the structures of the sets. How can we make that comparison?

    Let's draw number line graphs for the sets. The second graph will have one less dot than the first one because we do not know where to put a yet.

    The structure of the dots is almost identical. This is a reflection of the fact that the sets have equal σ-values. It also emphasises that the shift between sets does not impact the standard deviation.


    STEP: Determine the position and value of a
    [−1 point ⇒ 0 / 2 points left]

    We can locate the position of a because we know that the structure of the sets must match. Comparing the number line diagrams above, we can see that Set 2 is missing a value corresponding to 8 from Set 1. 2

    We can read the value of a from the number line now.

    The value of a is 14.


    Submit your answer as:
  2. The data below have a mean of 68.75 and a standard deviation of 2.22.

    65;66;67;70;70;70;71;71x¯=68.75σ=2.22

    If 65 is removed, would the value of σ increase, decrease, or stay the same?

    TIP: Do not calculate the new standard deviation. Use what you know about the concept of standard deviation.
    Answer: The value of σ would .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Is 65 within one σ of the mean value? Or is it outside of that boundary?


    STEP: Think about the distance of 65 from the mean
    [−1 point ⇒ 0 / 1 points left]

    To answer this question, we need to compare 65 to the boundary defined by σ around the mean. If we remove a number which is inside this boundary, σ should increase. That happens because we are removing a number with below-average spread and the spread will increase. But if we remove a number which is outside this boundary, σ should decrease. That happens because we are taking away a number which is far from the mean which reduces the overall spread.

    It is easiest to see this on a number line graph of the data values.

    The graph shows that the removed data value 65 is well outside the one σ boundary around x¯. So removing it from the data set will decrease the overall spread of the data set.

    If 65 is removed the value of σ will decrease.


    Submit your answer as:
  3. The following data set has ten values in total with a single unknown value, x:

    14;6;7;15;12;13;7;7;14;x

    The graph below shows the functions σ(x) and n¯(x) over a domain starting at x=0 and reaching an unspecified maximum value, xmax. The value x=p is marked on the horizontal axis.

    To approximately one decimal place, what is the value of p?

    Answer: p
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Do not try to calculate the answer. You can read the answer from the graph.


    STEP: Apply the concept of standard deviation from the mean
    [−1 point ⇒ 0 / 1 points left]

    We need to find a way to determine the value of p from the graph. Do you notice anything special about x=p?

    The only noteworthy thing about x=p is that it is where the standard deviation function has a minimum value. It turns out that we can use this fact to determine p!

    As x increases from 0, σ(x) decreases. This must be because the value of x is getting closer to the mean of the data set: that decreases the overall spread of the data set, pushing the value of σ down. But then, after x=p, σ(x) turns around and starts to increase. This means x is now getting further from the mean of the data, not closer, which increases the overall spread of the data. So at x=p (at the minimum value of σ(x)), x must be as close as possible to the mean.

    What is the closest x can get to the mean? It can be equal to the mean! So at x=p, x must be equal to the mean. This is shown on the graph below.

    The minimum value of σ(x) occurs when x is equal to the mean. The mean is labelled on the right axis, so that is where we need to read it off: p10.6.

    NOTE:

    It is possible to calculate the exact answer. We know that p is the value for which x is equal to the mean.

    x=x+(14+6+7+15+12+13+7+7+14)1010x=x+959x=95x=95910.6

    This answer is consistent with the estimation we found from the graph above.

    Check out the meaning of the fraction, 959: it is the sum of the nine known data values (the ten values without x) divided by 9. That is the mean of those nine values. This is really cool: it tells us that to get the smallest standard deviation possible for all ten numbers, x should be equal to the mean of the other nine numbers. That is deep: if x is equal to the mean of those nine numbers, then x itself adds nothing more to the standard deviation of those numbers. Maths is amazing - that is worth at least two exclamation marks!!

    The value of p is approximately 10.6.


    Submit your answer as:

ID is: 1292 Seed is: 6094

Calculating standard deviation

A group of 10 students count the number of playing cards they each have. This is the data they collect:

2;7;4;8;82;9;4;3;9

Calculate the standard deviation, σ, of the data set.

INSTRUCTION: Round your answer to 2 decimal places.
Answer: The standard deviation is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The formula for standard deviation is:

σ=(xix¯)2N

STEP: Write down the formula for the standard deviation and begin finding the required information
[−1 point ⇒ 3 / 4 points left]

The formula for standard deviation is as follows:

σ=(xix¯)2N

where:

  • N represents the number of data values.
  • x¯ represents the mean of the data set.
  • xi represents all the data values.
  • Σ indicates a sum, which means we need to add the quantities (xix¯)2. This means the numerator is made of many terms, not just one.

We can expand the numerator to show how each of the data values is involved in the calculation:

σ=(x1x¯)2+(x2x¯)2++(xNx¯)210

In the equation above, x1 represents the first data value. x2 represents the second data value. There is a term in the numerator for each and every data value.


STEP: Determine the values of N and the mean of the data set
[−1 point ⇒ 2 / 4 points left]

The standard deviation formula requires that we find the value of N:

N=number of data valuesN=10

We also need the mean of the data values:

x¯=the mean of the datax¯=2+7+4+8+8+2+9+4+3+910x¯=5.6

STEP: Substitute into the formula and evaluate
[−1 point ⇒ 1 / 4 points left]

Now we substitute the values into the formula and evaluate:

σ=(xix¯)2N=(xi5.6)210=(25.6)2+(75.6)2++(95.6)210

STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can continue with the calculation. Evaluate the values inside the brackets, square them, and add everything in the numerator together.

σ=(25.6)2+(75.6)2++(95.6)210=(3.6)2+(1.4)2++(3.4)210=(12.96)+(1.96)++(11.56)10=74.410

Finally, we can evalaute the fraction and then take the square root to get the final answer.

σ=74.410=7.44=2.72763...2.73
NOTE: Most calculators can find the standard deviation of a data set for you. It will be useful for you to learn how to do this with your calculator. However, it is also important to understand the formula for the standard deviation in detail. So being able to work out the complete calculation as shown above is a necessary step.

The standard deviation of the data, rounded to two decimal places, is: σ=2.73.


Submit your answer as:

ID is: 1292 Seed is: 9032

Calculating standard deviation

A group of 10 students count the number of playing cards they each have. This is the data they collect:

8;10;1;5;18;7;3;6;1

Calculate the standard deviation, σ, of the data set.

INSTRUCTION: Round your answer to 2 decimal places.
Answer: The standard deviation is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The formula for standard deviation is:

σ=(xix¯)2N

STEP: Write down the formula for the standard deviation and begin finding the required information
[−1 point ⇒ 3 / 4 points left]

The formula for standard deviation is as follows:

σ=(xix¯)2N

where:

  • N represents the number of data values.
  • x¯ represents the mean of the data set.
  • xi represents all the data values.
  • Σ indicates a sum, which means we need to add the quantities (xix¯)2. This means the numerator is made of many terms, not just one.

We can expand the numerator to show how each of the data values is involved in the calculation:

σ=(x1x¯)2+(x2x¯)2++(xNx¯)210

In the equation above, x1 represents the first data value. x2 represents the second data value. There is a term in the numerator for each and every data value.


STEP: Determine the values of N and the mean of the data set
[−1 point ⇒ 2 / 4 points left]

The standard deviation formula requires that we find the value of N:

N=number of data valuesN=10

We also need the mean of the data values:

x¯=the mean of the datax¯=8+10+1+5+1+8+7+3+6+110x¯=5

STEP: Substitute into the formula and evaluate
[−1 point ⇒ 1 / 4 points left]

Now we substitute the values into the formula and evaluate:

σ=(xix¯)2N=(xi5)210=(85)2+(105)2++(15)210

STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can continue with the calculation. Evaluate the values inside the brackets, square them, and add everything in the numerator together.

σ=(85)2+(105)2++(15)210=(3)2+(5)2++(4)210=(9)+(25)++(16)10=10010

Finally, we can evalaute the fraction and then take the square root to get the final answer.

σ=10010=10=3.16227...3.16
NOTE: Most calculators can find the standard deviation of a data set for you. It will be useful for you to learn how to do this with your calculator. However, it is also important to understand the formula for the standard deviation in detail. So being able to work out the complete calculation as shown above is a necessary step.

The standard deviation of the data, rounded to two decimal places, is: σ=3.16.


Submit your answer as:

ID is: 1292 Seed is: 8779

Calculating standard deviation

A group of 10 students count the number of marbles they each have. This is the data they collect:

9;8;6;4;55;8;1;6;2

Calculate the standard deviation, σ, of the data set.

INSTRUCTION: Round your answer to 2 decimal places.
Answer: The standard deviation is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The formula for standard deviation is:

σ=(xix¯)2N

STEP: Write down the formula for the standard deviation and begin finding the required information
[−1 point ⇒ 3 / 4 points left]

The formula for standard deviation is as follows:

σ=(xix¯)2N

where:

  • N represents the number of data values.
  • x¯ represents the mean of the data set.
  • xi represents all the data values.
  • Σ indicates a sum, which means we need to add the quantities (xix¯)2. This means the numerator is made of many terms, not just one.

We can expand the numerator to show how each of the data values is involved in the calculation:

σ=(x1x¯)2+(x2x¯)2++(xNx¯)210

In the equation above, x1 represents the first data value. x2 represents the second data value. There is a term in the numerator for each and every data value.


STEP: Determine the values of N and the mean of the data set
[−1 point ⇒ 2 / 4 points left]

The standard deviation formula requires that we find the value of N:

N=number of data valuesN=10

We also need the mean of the data values:

x¯=the mean of the datax¯=9+8+6+4+5+5+8+1+6+210x¯=5.4

STEP: Substitute into the formula and evaluate
[−1 point ⇒ 1 / 4 points left]

Now we substitute the values into the formula and evaluate:

σ=(xix¯)2N=(xi5.4)210=(95.4)2+(85.4)2++(25.4)210

STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can continue with the calculation. Evaluate the values inside the brackets, square them, and add everything in the numerator together.

σ=(95.4)2+(85.4)2++(25.4)210=(3.6)2+(2.6)2++(3.4)210=(12.96)+(6.76)++(11.56)10=60.410

Finally, we can evalaute the fraction and then take the square root to get the final answer.

σ=60.410=6.04=2.45764...2.46
NOTE: Most calculators can find the standard deviation of a data set for you. It will be useful for you to learn how to do this with your calculator. However, it is also important to understand the formula for the standard deviation in detail. So being able to work out the complete calculation as shown above is a necessary step.

The standard deviation of the data, rounded to two decimal places, is: σ=2.46.


Submit your answer as:

ID is: 3777 Seed is: 4830

Standard deviation and the spread of data

  1. Which of the data sets below would you expect to have the least variability as measured by the standard deviation, and why?

    Choose your answer from the options below.

    A Set 1, because it is the most evenly spread out.
    B Set 2, because it has the most values close to the middle.
    C Set 3, because it has the most values away from the middle
    D All three data sets would have the same standard deviation.
    Answer: The correct choice is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You want to find the set with the least standard deviation. That means you need to find the set which is most tightly bunched up near the mean (the middle).


    STEP: Compare the spread of the three sets
    [−1 point ⇒ 0 / 1 points left]

    We need to find the set with the least standard deviation. Standard deviation measures the spread of the data values around the mean. The arrows below show the distance of the data values from the mean for each set. (The mean must be in the centre because the data sets are all symmetric.)

    By comparing the arrows, we can see that Set 2 has the smallest standard deviation. Set 3 has the largest standard deviation. And Set 1 has a standard deviation value in between.

    NOTE:

    Here are two interesting facts about this question:

    • We do not know the data values for these sets. They could be in the range 0 up to 10 or from 500 up to 1,000. But that does not matter because the standard deviation only cares about how much the values are separated from the mean. The values themselves do not matter!
    • The formula for standard deviation includes (xx¯). This expression is the distance of each data value from the mean. So the arrows in the figure above correspond directly to that expression in the standard deviation formula. And the larger the arrows become, the larger the standard deviation becomes. This is how standard deviation formula measures the spread of the data from the mean!

    The correct option is Set 2, because it has the most values close to the middle..


    Submit your answer as:
  2. The histogram below summarises a set of 37 data values with a mean value of x¯309.92.

    Which of the numbers below is the best estimate for the standard deviation (rounded to two decimal places) of the data in the histogram?

    A 16
    B 0
    C 317.95
    D 4.01
    Answer:

    The best estimate for σ is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start by finding the range of the data. The standard deviation must be less than the range.


    STEP: Identify which value is most reasonable for the spread of the data
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation tells us how far we have to go away from the mean of a data set to surround the core of the data. For this histogram, the mean must be close to the centre because the histogram is nearly symmetric. The standard deviation must extend from the mean to include most of the data, but not all of it. On the histogram it must look something like this:

    NOTE: The above is an estimate! There is no calculation involved! We are only trying to approximate the standard deviation.

    We need to estimate the length of the arrows. Comparing to the x-axis, they are approximately 4 or 5 wide. So the value of σ must be close to that. How does this compare to the choices in the list?

    A ✘ 16
    The answer should be somewhat smaller than 16 because the standard deviation does not span the entire data set.
    B ✘ 0
    The answer cannot be zero because the data are not all equal to each other.
    C ✘ 317.95
    The spread of the data is much smaller than 317.95.
    D ✔ 4.01
    This is the best match for the figure above.

    Submit your answer as:

ID is: 3777 Seed is: 50

Standard deviation and the spread of data

  1. Which of the data sets below would you expect to have the least variability as measured by the standard deviation, and why?

    Choose your answer from the options below.

    A Set 1, because it is the most evenly spread out.
    B Set 2, because it has the most values close to the middle.
    C Set 3, because it has the most values away from the middle
    D All three data sets would have the same standard deviation.
    Answer: The correct choice is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You want to find the set with the least standard deviation. That means you need to find the set which is most tightly bunched up near the mean (the middle).


    STEP: Compare the spread of the three sets
    [−1 point ⇒ 0 / 1 points left]

    We need to find the set with the least standard deviation. Standard deviation measures the spread of the data values around the mean. The arrows below show the distance of the data values from the mean for each set. (The mean must be in the centre because the data sets are all symmetric.)

    By comparing the arrows, we can see that Set 2 has the smallest standard deviation. Set 3 has the largest standard deviation. And Set 1 has a standard deviation value in between.

    NOTE:

    Here are two interesting facts about this question:

    • We do not know the data values for these sets. They could be in the range 0 up to 10 or from 500 up to 1,000. But that does not matter because the standard deviation only cares about how much the values are separated from the mean. The values themselves do not matter!
    • The formula for standard deviation includes (xx¯). This expression is the distance of each data value from the mean. So the arrows in the figure above correspond directly to that expression in the standard deviation formula. And the larger the arrows become, the larger the standard deviation becomes. This is how standard deviation formula measures the spread of the data from the mean!

    The correct option is Set 2, because it has the most values close to the middle..


    Submit your answer as:
  2. The histogram below summarises a set of 47 data values with a mean value of x¯209.89.

    Which of the numbers below is the best estimate for the standard deviation (rounded to two decimal places) of the data in the histogram?

    A 12
    B 209.9
    C 0.56
    D 3.03
    Answer:

    The best estimate for σ is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start by finding the range of the data. The standard deviation must be less than the range.


    STEP: Identify which value is most reasonable for the spread of the data
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation tells us how far we have to go away from the mean of a data set to surround the core of the data. For this histogram, the mean must be close to the centre because the histogram is nearly symmetric. The standard deviation must extend from the mean to include most of the data, but not all of it. On the histogram it must look something like this:

    NOTE: The above is an estimate! There is no calculation involved! We are only trying to approximate the standard deviation.

    We need to estimate the length of the arrows. Comparing to the x-axis, they are approximately 3 or 4 wide. So the value of σ must be close to that. How does this compare to the choices in the list?

    A ✘ 12
    The answer should be somewhat smaller than 12 because the standard deviation does not span the entire data set.
    B ✘ 209.9
    The spread of the data is much smaller than 209.9.
    C ✘ 0.56
    This number is too small, because it implies the values are packed very close to the mean.
    D ✔ 3.03
    This is the best match for the figure above.

    Submit your answer as:

ID is: 3777 Seed is: 7658

Standard deviation and the spread of data

  1. Which of the data sets below would you expect to have the least variability as measured by the standard deviation, and why?

    Choose your answer from the options below.

    A Set 1, because it has the most values away from the middle
    B Set 2, because it is the most evenly spread out.
    C Set 3, because it has the most values close to the middle.
    D All three data sets would have the same standard deviation.
    Answer: The correct choice is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You want to find the set with the least standard deviation. That means you need to find the set which is most tightly bunched up near the mean (the middle).


    STEP: Compare the spread of the three sets
    [−1 point ⇒ 0 / 1 points left]

    We need to find the set with the least standard deviation. Standard deviation measures the spread of the data values around the mean. The arrows below show the distance of the data values from the mean for each set. (The mean must be in the centre because the data sets are all symmetric.)

    By comparing the arrows, we can see that Set 3 has the smallest standard deviation. Set 1 has the largest standard deviation. And Set 2 has a standard deviation value in between.

    NOTE:

    Here are two interesting facts about this question:

    • We do not know the data values for these sets. They could be in the range 0 up to 10 or from 500 up to 1,000. But that does not matter because the standard deviation only cares about how much the values are separated from the mean. The values themselves do not matter!
    • The formula for standard deviation includes (xx¯). This expression is the distance of each data value from the mean. So the arrows in the figure above correspond directly to that expression in the standard deviation formula. And the larger the arrows become, the larger the standard deviation becomes. This is how standard deviation formula measures the spread of the data from the mean!

    The correct option is Set 3, because it has the most values close to the middle..


    Submit your answer as:
  2. The histogram below summarises a set of 47 data values with a mean value of x¯419.55.

    Which of the numbers below is the best estimate for the standard deviation (rounded to two decimal places) of the data in the histogram?

    A 419.6
    B 10.29
    C 0
    D 40
    Answer:

    The best estimate for σ is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start by finding the range of the data. The standard deviation must be less than the range.


    STEP: Identify which value is most reasonable for the spread of the data
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation tells us how far we have to go away from the mean of a data set to surround the core of the data. For this histogram, the mean must be close to the centre because the histogram is nearly symmetric. The standard deviation must extend from the mean to include most of the data, but not all of it. On the histogram it must look something like this:

    NOTE: The above is an estimate! There is no calculation involved! We are only trying to approximate the standard deviation.

    We need to estimate the length of the arrows. Comparing to the x-axis, they are approximately 10 or 11 wide. So the value of σ must be close to that. How does this compare to the choices in the list?

    A ✘ 419.6
    The spread of the data is much smaller than 419.6.
    B ✔ 10.29
    This is the best match for the figure above.
    C ✘ 0
    The answer cannot be zero because the data are not all equal to each other.
    D ✘ 40
    The answer should be somewhat smaller than 40 because the standard deviation does not span the entire data set.

    Submit your answer as:

ID is: 3780 Seed is: 325

What does standard deviation measure?

Answer the two questions below about the standard deviation of a data set.

  1. What does the standard deviation measure?

    Standard deviation measures...

    A which number is the most common.
    B how many values there are in a data set.
    C how high the numbers go.
    D how much a data set is spread out.
    Answer:

    The correct answer is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The standard deviation measures how close together or far apart the numbers in a set of data are.


    STEP: Identify the statement which describes standard deviation
    [−1 point ⇒ 0 / 1 points left]

    This question is about the meaning of standard deviation. The pictures below represent the concept of standard deviation.

    Low standard deviation

    High standard deviation

    The first picture above represents what a small standard deviation looks like: the runners are all on the starting line. But as the race continues, the runners spread out. So the second picture represents a larger standard deviation than the first picture.

    The standard deviation summarises how values in a data set are spread out from each other. So the correct choice is D.


    Submit your answer as:
  2. Given the two graphs below, which one shows a larger standard deviation?

    Answer:

    Graph has the larger standard deviation.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think about the pictures of the race in Question 1: which picture shows the runners more spread out? We need to identify the same thing about the histograms.


    STEP: Compare the widths of the graphs
    [−1 point ⇒ 0 / 1 points left]

    We need to use the concept from Question 1. The graph with the larger standard deviation is the one which shows spread out data values. Think about the pictures of the race in Question 1: which picture shows the runners more spread out? We need to identify the same thing about the histograms.

    We can compare these histograms to the pictures in Question 1. The first histogram resembles the image of the starting line and the second histogram resembles the image of the finishing line. So Graph A has a smaller standard deviation than Graph B.

    The graph with the larger standard deviation is Graph B.


    Submit your answer as:

ID is: 3780 Seed is: 2790

What does standard deviation measure?

Answer the two questions below about the standard deviation of a data set.

  1. What does the standard deviation measure?

    Standard deviation measures...

    A how many values there are in a data set.
    B how much a data set is spread out.
    C how high the numbers go.
    D which number is the most important.
    Answer:

    The correct answer is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The standard deviation measures how close together or far apart the numbers in a set of data are.


    STEP: Identify the statement which describes standard deviation
    [−1 point ⇒ 0 / 1 points left]

    This question is about the meaning of standard deviation. The pictures below represent the concept of standard deviation.

    Low standard deviation

    High standard deviation

    The first picture above represents what a small standard deviation looks like: the runners are all on the starting line. But as the race continues, the runners spread out. So the second picture represents a larger standard deviation than the first picture.

    The standard deviation summarises how values in a data set are spread out from each other. So the correct choice is B.


    Submit your answer as:
  2. Given the two graphs below, which one shows a larger standard deviation?

    Answer:

    Graph has the larger standard deviation.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think about the pictures of the race in Question 1: which picture shows the runners more spread out? We need to identify the same thing about the histograms.


    STEP: Compare the widths of the graphs
    [−1 point ⇒ 0 / 1 points left]

    We need to use the concept from Question 1. The graph with the larger standard deviation is the one which shows spread out data values. Think about the pictures of the race in Question 1: which picture shows the runners more spread out? We need to identify the same thing about the histograms.

    We can compare these histograms to the pictures in Question 1. The first histogram resembles the image of the starting line and the second histogram resembles the image of the finishing line. So Graph A has a smaller standard deviation than Graph B.

    The graph with the larger standard deviation is Graph B.


    Submit your answer as:

ID is: 3780 Seed is: 2044

What does standard deviation measure?

Answer the two questions below about the standard deviation of a data set.

  1. What does the standard deviation measure?

    Standard deviation measures...

    A how far apart the minimum and maximum values are.
    B which number is the most common.
    C how much a data set is spread out.
    D how many values there are in a data set.
    Answer:

    The correct answer is choice .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The standard deviation measures how close together or far apart the numbers in a set of data are.


    STEP: Identify the statement which describes standard deviation
    [−1 point ⇒ 0 / 1 points left]

    This question is about the meaning of standard deviation. The pictures below represent the concept of standard deviation.

    Low standard deviation

    High standard deviation

    During a rugby match the players can be quite spread out. But during the scrum most of the players bunch up in a clump. The scrum shows the concept of a small standard deviation, while the second photo represents a larger standard deviation.

    The standard deviation summarises how values in a data set are spread out from each other. So the correct choice is C.


    Submit your answer as:
  2. Given the two graphs below, which one shows a smaller standard deviation?

    Answer:

    Graph has the smaller standard deviation.

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think about the rugby pictures in Question 1: which picture shows the rugby players less spread out? We need to identify the same thing about the histograms.


    STEP: Compare the widths of the graphs
    [−1 point ⇒ 0 / 1 points left]

    We need to use the concept from Question 1. The graph with the smaller standard deviation is the one which shows tightly clustered data values. Think about the rugby pictures in Question 1: which picture shows the rugby players less spread out? We need to identify the same thing about the histograms.

    We can compare these histograms to the pictures in Question 1. The first histogram resembles the image of the scrum and the second histogram resembles the image of the rugby players spread out on the pitch. So Graph A has a smaller standard deviation than Graph B.

    The graph with the smaller standard deviation is Graph A.


    Submit your answer as:

ID is: 3778 Seed is: 3549

Estimating the standard deviation from a graph

The graph below shows a perfectly symmetric data set.

  1. Estimate the mean of the data set.
  2. Estimate the standard deviation of the data set.
Answer:
  1. The mean is approximately .
  2. The standard deviation is approximately .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The mean of the values must be in the middle of the curve because the curve is symmetric. After you find the mean, you can use it to find the standard deviation.


STEP: Estimate the mean value
[−1 point ⇒ 1 / 2 points left]

The mean of a data set summarises the data by telling us one number around which the set is balanced. If the data set is symmetric, like the one in this graph, the mean will be exactly in the middle of the set.

We can estimate the mean by reading it from the graph.

The mean of this data set is approximately 25.


STEP: Estimate the standard deviation
[−1 point ⇒ 0 / 2 points left]

The standard deviation measures the spread of a data set. It does this by showing us a zone which includes the core of the data set. If the data are spread far apart, that zone must be wider, making the standard deviation larger.

The arrows in the graph below show the standard deviation. The zone around the mean is shaded below. It reaches from the mean down to 17 and from the mean up to 33.

We want to know how wide each half of that zone is because the standard deviation is equal to the length of each arrow. We can work out the length of the arrows in two ways:

standard deviation=upper limitmean33258ORstandard deviation=meanlower limit25178

The standard deviation of this data set is approximately 8.


Submit your answer as: and

ID is: 3778 Seed is: 3655

Estimating the standard deviation from a graph

The graph below shows a perfectly symmetric data set.

  1. Estimate the mean of the data set.
  2. Estimate the standard deviation of the data set.
Answer:
  1. The mean is approximately .
  2. The standard deviation is approximately .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The mean of the values must be in the middle of the curve because the curve is symmetric. After you find the mean, you can use it to find the standard deviation.


STEP: Estimate the mean value
[−1 point ⇒ 1 / 2 points left]

The mean of a data set summarises the data by telling us one number around which the set is balanced. If the data set is symmetric, like the one in this graph, the mean will be exactly in the middle of the set.

We can estimate the mean by reading it from the graph.

The mean of this data set is approximately 120.


STEP: Estimate the standard deviation
[−1 point ⇒ 0 / 2 points left]

The standard deviation measures the spread of a data set. It does this by showing us a zone which includes the core of the data set. If the data are spread far apart, that zone must be wider, making the standard deviation larger.

The arrows in the graph below show the standard deviation. The zone around the mean is shaded below. It reaches from the mean down to 110 and from the mean up to 130.

We want to know how wide each half of that zone is because the standard deviation is equal to the length of each arrow. We can work out the length of the arrows in two ways:

standard deviation=upper limitmean13012010ORstandard deviation=meanlower limit12011010

The standard deviation of this data set is approximately 10.


Submit your answer as: and

ID is: 3778 Seed is: 1513

Estimating the standard deviation from a graph

The graph below shows a perfectly symmetric data set.

  1. Estimate the mean of the data set.
  2. Estimate the standard deviation of the data set.
Answer:
  1. The mean is approximately .
  2. The standard deviation is approximately .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The mean of the values must be in the middle of the curve because the curve is symmetric. After you find the mean, you can use it to find the standard deviation.


STEP: Estimate the mean value
[−1 point ⇒ 1 / 2 points left]

The mean of a data set summarises the data by telling us one number around which the set is balanced. If the data set is symmetric, like the one in this graph, the mean will be exactly in the middle of the set.

We can estimate the mean by reading it from the graph.

The mean of this data set is approximately 34.


STEP: Estimate the standard deviation
[−1 point ⇒ 0 / 2 points left]

The standard deviation measures the spread of a data set. It does this by showing us a zone which includes the core of the data set. If the data are spread far apart, that zone must be wider, making the standard deviation larger.

The arrows in the graph below show the standard deviation. The zone around the mean is shaded below. It reaches from the mean down to 29 and from the mean up to 39.

We want to know how wide each half of that zone is because the standard deviation is equal to the length of each arrow. We can work out the length of the arrows in two ways:

standard deviation=upper limitmean39345ORstandard deviation=meanlower limit34295

The standard deviation of this data set is approximately 5.


Submit your answer as: and

ID is: 3783 Seed is: 9779

Grouping values using sigma

The information below shows a set of 18 data points. The mean of the data points is x¯=32.5 and the standard deviation is σ2.1.

27;29;30;31;31;32;32;33;33;33;33;34;34;34;34;35;35;35x¯=32.5σ2.1
  1. Which of the values below are outside of one standard deviation of the mean?

    A 27
    B 29
    C 30
    D 35
    INSTRUCTION: Identify all correct values.
    Answer: The correct values are .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start by calculating 32.52.1 and 32.5+2.1.


    STEP: Compare the numbers in the table to the core of the data set
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation identifies a core of values in a data set which is centred on the mean. We find the boundaries of this core using the mean and the standard deviation:

    lower boundary=32.52.1=30.4upper boundary=32.5+2.1=34.6

    If we graph the data and these boundaries on a number line, it looks like this:

    The shaded area above is the zone within one standard deviation of the mean. Any values in the data set from 30.4 up to 34.6 are within one standard deviation of the mean. Values which are less than 30.4 or greater than 34.6 are more than one standard deviation away from the mean.

    Which of the numbers in the table are outside of one standard deviation of the mean?

    A 27
    B 29
    C 30
    D 35

    In this case, all four of them are. They are shown with the green smiley faces below.

    The correct choice is All four values.


    Submit your answer as:
  2. For the data in this question, there are 6 data points outside of one standard deviation of the mean. How many points are within the range of the data?

    Answer:

    There are points within the range.

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The range of a set is the difference between the minimum and maximum values. How many values are within that range?


    STEP: Use the definition of the range
    [−1 point ⇒ 0 / 1 points left]

    The range of the data set is the total width of the data set. It is the distance from the minimum value to the maximum value. For the data in this question, the minimum value is 27 and the maximum is 35. So the range is:

    range=3527=8

    So the width, from the smallest number in the set to the biggest, is 8. By definition, every value in the set is within this range. That means all 18 data points in the set must be within the range.

    NOTE: The range and the standard deviation are both measures of dispersion: they both summarise the spread of a data set. But the range does not care about any details within the data. In fact, the range comes from only two of the data values: the minimum and the maximum. In contrast, the standard deviation compares every value in the set to the mean! Hence the standard deviation usually gives us a much better idea of the structure of the data than the range can.

    The number of points within the range is 18.


    Submit your answer as:

ID is: 3783 Seed is: 6839

Grouping values using sigma

The information below shows a set of 15 data points. The mean of the data points is x¯=25.8 and the standard deviation is σ2.1.

22;23;24;24;25;25;25;25;26;27;27;28;28;28;30x¯=25.8σ2.1
  1. Which of the values below are within one standard deviation of the mean?

    A 22
    B 23
    C 28
    D 30
    INSTRUCTION: Identify all correct values.
    Answer: The correct values are .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start by calculating 25.82.1 and 25.8+2.1.


    STEP: Compare the numbers in the table to the core of the data set
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation identifies a core of values in a data set which is centred on the mean. We find the boundaries of this core using the mean and the standard deviation:

    lower boundary=25.82.1=23.7upper boundary=25.8+2.1=27.9

    If we graph the data and these boundaries on a number line, it looks like this:

    The shaded area above is the zone within one standard deviation of the mean. Any values in the data set from 23.7 up to 27.9 are within one standard deviation of the mean. Values which are less than 23.7 or greater than 27.9 are more than one standard deviation away from the mean.

    Which of the numbers in the table are within one standard deviation of the mean?

    A 22
    B 23
    C 28
    D 30

    In this case, none of them are. They are shown with the red frown faces below.

    The correct choice is None of the above.


    Submit your answer as:
  2. For the data in this question, there are 9 data points within one standard deviation of the mean. How many points are within the range of the data?

    Answer:

    There are points within the range.

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The range of a set is the difference between the minimum and maximum values. How many values are within that range?


    STEP: Use the definition of the range
    [−1 point ⇒ 0 / 1 points left]

    The range of the data set is the total width of the data set. It is the distance from the minimum value to the maximum value. For the data in this question, the minimum value is 22 and the maximum is 30. So the range is:

    range=3022=8

    So the width, from the smallest number in the set to the biggest, is 8. By definition, every value in the set is within this range. That means all 15 data points in the set must be within the range.

    NOTE: The range and the standard deviation are both measures of dispersion: they both summarise the spread of a data set. But the range does not care about any details within the data. In fact, the range comes from only two of the data values: the minimum and the maximum. In contrast, the standard deviation compares every value in the set to the mean! Hence the standard deviation usually gives us a much better idea of the structure of the data than the range can.

    The number of points within the range is 15.


    Submit your answer as:

ID is: 3783 Seed is: 4677

Grouping values using sigma

The information below shows a set of 20 data points. The mean of the data points is x¯=27.3 and the standard deviation is σ2.6.

23;23;24;25;25;25;26;26;27;27;27;28;28;29;29;30;31;31;31;31x¯=27.3σ2.6
  1. Which of the values below are within one standard deviation of the mean?

    A 25
    B 27
    C 28
    D 29
    INSTRUCTION: Identify all correct values.
    Answer: The correct values are .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Start by calculating 27.32.6 and 27.3+2.6.


    STEP: Compare the numbers in the table to the core of the data set
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation identifies a core of values in a data set which is centred on the mean. We find the boundaries of this core using the mean and the standard deviation:

    lower boundary=27.32.6=24.7upper boundary=27.3+2.6=29.9

    If we graph the data and these boundaries on a number line, it looks like this:

    The shaded area above is the zone within one standard deviation of the mean. Any values in the data set from 24.7 up to 29.9 are within one standard deviation of the mean. Values which are less than 24.7 or greater than 29.9 are more than one standard deviation away from the mean.

    Which of the numbers in the table are within one standard deviation of the mean?

    A 25
    B 27
    C 28
    D 29

    In this case, all four of them are. They are shown with the green smiley faces below.

    The correct choice is All four values.


    Submit your answer as:
  2. For the data in this question, there are 12 data points within one standard deviation of the mean. How many points are within the range of the data?

    Answer:

    There are points within the range.

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The range of a set is the difference between the minimum and maximum values. How many values are within that range?


    STEP: Use the definition of the range
    [−1 point ⇒ 0 / 1 points left]

    The range of the data set is the total width of the data set. It is the distance from the minimum value to the maximum value. For the data in this question, the minimum value is 23 and the maximum is 31. So the range is:

    range=3123=8

    So the width, from the smallest number in the set to the biggest, is 8. By definition, every value in the set is within this range. That means all 20 data points in the set must be within the range.

    NOTE: The range and the standard deviation are both measures of dispersion: they both summarise the spread of a data set. But the range does not care about any details within the data. In fact, the range comes from only two of the data values: the minimum and the maximum. In contrast, the standard deviation compares every value in the set to the mean! Hence the standard deviation usually gives us a much better idea of the structure of the data than the range can.

    The number of points within the range is 20.


    Submit your answer as:

ID is: 3776 Seed is: 3006

Features of a data set graph

The graph below represents a perfectly symmetric data set. The labels A, B, C, and D indicate features on this graph.

Which of the labels corresponds to the standard deviation of the data?

Answer:

The standard deviation corresponds to Label .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
  • The mean represents the centre of the data.
  • The standard deviation represents how much the data are spread out.
  • The minimum and maximum data values are at the beginning and end of the graph.

STEP: Identify the standard deviation on the graph
[−1 point ⇒ 0 / 1 points left]

The figure below shows how the standard deviation is related to other key features of a data set.

The picture above shows the following key facts about the standard deviation:

  • The standard deviation is attached to the mean value.
  • The standard deviation usually does not surround all the data values. You can see that the minimum and maximum data values are both more than one standard deviation from the mean.

The standard deviation of the data corresponds to Label C.


Submit your answer as:

ID is: 3776 Seed is: 5999

Features of a data set graph

The graph below represents a perfectly symmetric data set. The labels A, B, C, and D indicate features on this graph.

Which of the labels corresponds to the standard deviation of the data?

Answer:

The standard deviation corresponds to Label .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
  • The mean represents the centre of the data.
  • The standard deviation represents how much the data are spread out.
  • The minimum and maximum data values are at the beginning and end of the graph.

STEP: Identify the standard deviation on the graph
[−1 point ⇒ 0 / 1 points left]

The figure below shows how the standard deviation is related to other key features of a data set.

The picture above shows the following key facts about the standard deviation:

  • The standard deviation is attached to the mean value.
  • The standard deviation usually does not surround all the data values. You can see that the minimum and maximum data values are both more than one standard deviation from the mean.

The standard deviation of the data corresponds to Label A.


Submit your answer as:

ID is: 3776 Seed is: 3572

Features of a data set graph

The graph below represents a perfectly symmetric data set. The labels A, B, C, and D indicate features on this graph.

Which of the labels corresponds to the mean value of the data?

Answer:

The mean value corresponds to Label .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
  • The mean represents the centre of the data.
  • The standard deviation represents how much the data are spread out.
  • The minimum and maximum data values are at the beginning and end of the graph.

STEP: Identify the mean value on the graph
[−1 point ⇒ 0 / 1 points left]

The figure below shows how the standard deviation is related to other key features of a data set.

The picture above shows the following key facts about the standard deviation:

  • The standard deviation is attached to the mean value.
  • The standard deviation usually does not surround all the data values. You can see that the minimum and maximum data values are both more than one standard deviation from the mean.

The mean value of the data corresponds to Label C.


Submit your answer as:

ID is: 3775 Seed is: 9733

What does σ=0 mean?

  1. If the standard deviation of a data set is equal to zero, which of the statements below must be true about the data?

    A The data values are equal to each other.
    B The data set includes positive and negative values with zero in the middle.
    C The data values are all equal to zero.
    D It is impossible to know exactly what it means without the data.
    Answer:

    The correct statement for σ=0 is .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation is a measure of the spread of data. If the standard deviation is zero, then the data are not spread out at all!


    STEP: Identify the description of a data set with σ=0
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation is a measure of how much the values in a data set are spread out from the mean of the data. In other words, standard deviation tells us how tightly the numbers are clustered around the mean.

    For example, the first histogram below represents data with a standard deviation of <number>4.02</number>. The second histogram represents data with a standard deviation of <number>1.91</number>. You can see that the first histogram is much more spread out (it is wider) and the second histogram shows values bunched tightly around the mean.

    A smaller standard deviation means that the numbers are tightly bunched up around the mean like people in a minibus taxi. But a large standard deviation means that the numbers are more spread out, like clouds in a mostly sunny sky.

    If the standard deviation is zero, that means there is no spread in the data. In other words, all the data values must be equal to each other.

    The correct choice is the data values are equal to each other., which is Choice A.


    Submit your answer as:
  2. The figure below shows a set of 8 values graphed on a number line. The mean of the data is x¯=8.

    Would you expect the standard deviation of these numbers to be less than zero, equal to zero, between zero and one, or greater than one? Select your answer using the list below.

    TIP: Do not calculate the standard deviation. Instead you should consider how the data values are spread out.
    Answer:

    The standard deviation must be .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think about the meaning of standard deviation. If the values are all close together the standard deviation must be small. But if the values are far apart from each other the standard deviation must be large.


    STEP: Look at the spread of the values
    [−1 point ⇒ 0 / 1 points left]

    In Question 1 we had an explanation of what standard deviation measures. For this question we need to decide if the standard deviation of the numbers given is equal to zero, greater than one, and so on.

    The numbers are sprinkled over a fairly wide range (from 3 up to 12). The standard deviation must be larger enough to represent that spread.

    The shaded block shows a zone around the mean which includes the core of the data values. This core includes most, but not all, of the values. In this case, the width of this zone is definitely bigger than one. Here are some guidelines about standard deviation:

    • If all the data values are equal to each other the standard deviation must be zero. This is because the values are not spread out at all.
    • The standard deviation usually surrounds most, but not all, of the data values. For example, you can see in the figure above that 3 and 12 are both outside the standard deviation.
    • The standard deviation is never negative.

    For the data in this question, the standard deviation must be greater than 1.


    Submit your answer as:

ID is: 3775 Seed is: 8284

What does σ=0 mean?

  1. If the standard deviation of a data set is equal to zero, which of the statements below must be true about the data?

    A The data set includes positive and negative values with zero in the middle.
    B The data values are equal to each other.
    C The data values are all equal to zero.
    D It is impossible to know exactly what it means without the data.
    Answer:

    The correct statement for σ=0 is .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation is a measure of the spread of data. If the standard deviation is zero, then the data are not spread out at all!


    STEP: Identify the description of a data set with σ=0
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation is a measure of how much the values in a data set are spread out from the mean of the data. In other words, standard deviation tells us how tightly the numbers are clustered around the mean.

    For example, the first histogram below represents data with a standard deviation of <number>4.07</number>. The second histogram represents data with a standard deviation of <number>2.02</number>. You can see that the first histogram is much more spread out (it is wider) and the second histogram shows values bunched tightly around the mean.

    A smaller standard deviation means that the numbers are tightly bunched up around the mean like people in a minibus taxi. But a large standard deviation means that the numbers are more spread out, like clouds in a mostly sunny sky.

    If the standard deviation is zero, that means there is no spread in the data. In other words, all the data values must be equal to each other.

    The correct choice is the data values are equal to each other., which is Choice B.


    Submit your answer as:
  2. The figure below shows a set of 8 values graphed on a number line. The mean of the data is x¯=8.25.

    Would you expect the standard deviation of these numbers to be less than zero, equal to zero, between zero and one, or greater than one? Select your answer using the list below.

    TIP: Do not calculate the standard deviation. Instead you should consider how the data values are spread out.
    Answer:

    The standard deviation must be .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think about the meaning of standard deviation. If the values are all close together the standard deviation must be small. But if the values are far apart from each other the standard deviation must be large.


    STEP: Look at the spread of the values
    [−1 point ⇒ 0 / 1 points left]

    In Question 1 we had an explanation of what standard deviation measures. For this question we need to decide if the standard deviation of the numbers given is equal to zero, greater than one, and so on.

    The numbers are all clustered in a small range from 7 up to 9. So the standard deviation must be bigger than zero but it must also be very small because the numbers are packed into such a small range.

    The shaded block shows a zone around the mean which includes the core of the data values. This core includes most, but not all, of the values. In this case, the width of this zone is less than one. Here are some guidelines about standard deviation:

    • If all the data values are equal to each other the standard deviation must be zero. This is because the values are not spread out at all.
    • The standard deviation usually surrounds most, but not all, of the data values. For example, you can see in the figure above that 7 and 7 are both outside the standard deviation.
    • The standard deviation is never negative.

    For the data in this question, the standard deviation must be between 0 and 1.


    Submit your answer as:

ID is: 3775 Seed is: 9543

What does σ=0 mean?

  1. If the standard deviation of a data set is equal to zero, which of the statements below must be true about the data?

    A The data values are equal to each other.
    B The data set includes positive and negative values with zero in the middle.
    C The data values are all equal to zero.
    D It is impossible to know exactly what it means without the data.
    Answer:

    The correct statement for σ=0 is .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Standard deviation is a measure of the spread of data. If the standard deviation is zero, then the data are not spread out at all!


    STEP: Identify the description of a data set with σ=0
    [−1 point ⇒ 0 / 1 points left]

    The standard deviation is a measure of how much the values in a data set are spread out from the mean of the data. In other words, standard deviation tells us how tightly the numbers are clustered around the mean.

    For example, the first histogram below represents data with a standard deviation of <number>4.07</number>. The second histogram represents data with a standard deviation of <number>1.99</number>. You can see that the first histogram is much more spread out (it is wider) and the second histogram shows values bunched tightly around the mean.

    A smaller standard deviation means that the numbers are tightly bunched up around the mean like people in a minibus taxi. But a large standard deviation means that the numbers are more spread out, like clouds in a mostly sunny sky.

    If the standard deviation is zero, that means there is no spread in the data. In other words, all the data values must be equal to each other.

    The correct choice is the data values are equal to each other., which is Choice A.


    Submit your answer as:
  2. The figure below shows a set of 6 values graphed on a number line. The mean of the data is x¯=7.83.

    Would you expect the standard deviation of these numbers to be less than zero, equal to zero, between zero and one, or greater than one? Select your answer using the list below.

    TIP: Do not calculate the standard deviation. Instead you should consider how the data values are spread out.
    Answer:

    The standard deviation must be .

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think about the meaning of standard deviation. If the values are all close together the standard deviation must be small. But if the values are far apart from each other the standard deviation must be large.


    STEP: Look at the spread of the values
    [−1 point ⇒ 0 / 1 points left]

    In Question 1 we had an explanation of what standard deviation measures. For this question we need to decide if the standard deviation of the numbers given is equal to zero, greater than one, and so on.

    The numbers are sprinkled over a fairly wide range (from 3 up to 12). The standard deviation must be larger enough to represent that spread.

    The shaded block shows a zone around the mean which includes the core of the data values. This core includes most, but not all, of the values. In this case, the width of this zone is definitely bigger than one. Here are some guidelines about standard deviation:

    • If all the data values are equal to each other the standard deviation must be zero. This is because the values are not spread out at all.
    • The standard deviation usually surrounds most, but not all, of the data values. For example, you can see in the figure above that 3 and 12 are both outside the standard deviation.
    • The standard deviation is never negative.

    For the data in this question, the standard deviation must be greater than 1.


    Submit your answer as:

ID is: 3796 Seed is: 4244

The meaning of standard deviation

Income inequality is about how much income differs between people in a society. A major goal in South Africa has been to reduce income inequality. This means people want to reduce the gap between rich people and poor people.

  1. Suppose someone collects data about the monthly income for thousands of people in South Africa and compares that data to similar data from ten years earlier. If the goal is to reduce income inequality, would you expect the data to show an increasing σ-value or a decreasing σ-value, and why?

    A σ should increase because incomes will get further apart.
    B σ should decrease because incomes will get closer together.
    C σ should increase because more people will get jobs.
    D Standard deviation should not change (it will not increase or decrease).
    Answer: The correct choice is option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the standard deviation decreases, that means the incomes are growing closer together. If the standard deviation increases, that means the incomes are getting further apart. Do either of those options make more sense?


    STEP: Consider the goal, and how it would appear in the data
    [−1 point ⇒ 0 / 1 points left]

    The question states that someone collects data about the monthly income for thousands of people in South Africa and compares that data to similar data from ten years earlier.

    Higher income inequality indicates a large gap between the rich and the poor. In other words, the incomes are spread far apart from each other. To reduce income inequality, the income values must get closer together. And that means the standard deviation should decrease.

    To reach the stated goal, σ should decrease because incomes will get closer together, which is choice B.


    Submit your answer as:
  2. The numbers below are a small data set showing monthly incomes for 10 people in South Africa (in rands).

    5,200;9,200;6,700;6,100;9,100;17,920;26,820;9,500;6,000;7,300

    Calculate the standard deviation of these values.

    INSTRUCTIONS:
    • Use may use a calculator.
    • Round your answer to two decimal places, if necessary.
    Answer: σ N=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by entering the numbers into your calculator. If you do not have a calculator, you will have to start by calculating the mean which you need for the formula:

    σ=Σ(xix¯)2N

    STEP: Enter the data into your calculator
    [−1 point ⇒ 1 / 2 points left]

    Most calculators will do statistics for you. If a question does not state that you cannot use your calculator, it is the best option. The first step will be to type in the data set, and then you can get statistical values like the mean and the standard deviation.

    If you are not sure how to use your calculator to calculate statisitcs, try these options:

    • search for a video about calculating statistics on your calculator
    • ask a classmate who might know
    • check with your teacher
    TIP: After you finish typing the data values into your calculator, check to see if the number of data values is correct. It is easy to skip a number or type a value twice. You can ask you calculator for the number of values you typed in and the check that it matches the total number of values. For this data set the number should be N=10.

    STEP: Ask your calculator for the standard deviation
    [−1 point ⇒ 0 / 2 points left]

    Now we can ask the calculator to do the serious work for us. Again, you might need to find information about how to do this. (If you do not have a calculator, or are not able to use it for these calculations, you will have to do the calculation yourself, using the formula below.) Most calculators have two versions of the standard deviation. Be sure to chose the correct one, which includes N and not N1.

    σ=Σ(xix¯)2N=6,467.56708...N=6,467.57
    TIP:

    You should check whether or not your answer is reasonable. Even though the calculator will not make a calculation error for σ, if you made any mistakes typing the data into the calculator the answer will be wrong. Possible data entry mistakes include:

    • typing a number incorrectly
    • leaving out a number
    • accidentally typing a number more than once

    One way to check if your answer is reasonable is by comparing the σ-value to the mean. (You can also ask your calculator for the mean.) Make sure that the 10 data values are a reasonable match for this mean and the sigma value that you get. In this case the mean is:

    x¯=N=10,384

    Rounded to two decimal places, the standard deviation is σ N= 6,467.57.


    Submit your answer as:

ID is: 3796 Seed is: 6066

The meaning of standard deviation

HIV damages a person's immune system by reducing the number of special blood cells in a person's blood. These special blood cells are called CD4 cells. People who are HIV-negative (those who do not have HIV) have CD4 counts spread fairly evenly from 500 up to 1,500. People who are HIV-positive can expect that their CD4 count will decrease below 500. If a person's CD4 count is 200 or less they are diagnosed with AIDS.

  1. Antiretroviral therapy is very effective for maintaining a CD4 count above 500 for people with HIV. Suppose someone collects the CD4 cell counts for thousands of people in South Africa (HIV-positive and HIV-negative) and compares that data to similar data taken ten years earlier. If the goal is to reduce the number of people with a CD4 count below 500, would you expect the data to show an increasing σ-value or a decreasing σ-value, and why?

    A σ should increase because there will be more low CD4 count values.
    B σ should decrease because there will be fewer low CD4 count values.
    C σ should increase because the average CD4 count will go up.
    D Standard deviation should not change (it will not increase or decrease).
    Answer: The correct choice is option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the standard deviation decreases, that means the CD4 counts are growing closer together. If the standard deviation increases, that means the CD4 counts are getting further apart. Do either of those options make more sense?


    STEP: Consider the goal, and how it would appear in the data
    [−1 point ⇒ 0 / 1 points left]

    The question states that oviral therapy is very effective for maintaining a CD4 count above 500 for people with HIV. Suppose someone collects the CD4 cell counts for thousands of people in South Africa (HIV-positive and HIV-negative) and compares that data to similar data taken ten years earlier.

    With a large population of HIV-positive people, we could expect many data values below 500 because HIV decreases the number of CD4 cells in a person's blood. In that case the values will range from very small (less than 200) up to about 1,500. But if we reach the stated goal, there will be fewer people with a CD4 count below 500 so most of the data values would be between 500 and the maximum of 1,500. This means the data set would have to get a bit less spread out because the low CD4 count values are increasing into the normal range. That means the standard deviation should decrease.

    To reach the stated goal, σ should decrease because there will be fewer low CD4 count values, which is choice B.


    Submit your answer as:
  2. The numbers below are a small data set showing CD4 counts for 10 people taken at random (number of cells).

    1,290;240;1,270;1,060;470;1,240;600;1,100;1,110;1,460

    Calculate the standard deviation of these values.

    INSTRUCTIONS:
    • Use may use a calculator.
    • Round your answer to two decimal places, if necessary.
    Answer: σ cells
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by entering the numbers into your calculator. If you do not have a calculator, you will have to start by calculating the mean which you need for the formula:

    σ=Σ(xix¯)2N

    STEP: Enter the data into your calculator
    [−1 point ⇒ 1 / 2 points left]

    Most calculators will do statistics for you. If a question does not state that you cannot use your calculator, it is the best option. The first step will be to type in the data set, and then you can get statistical values like the mean and the standard deviation.

    If you are not sure how to use your calculator to calculate statisitcs, try these options:

    • search for a video about calculating statistics on your calculator
    • ask a classmate who might know
    • check with your teacher
    TIP: After you finish typing the data values into your calculator, check to see if the number of data values is correct. It is easy to skip a number or type a value twice. You can ask you calculator for the number of values you typed in and the check that it matches the total number of values. For this data set the number should be N=10.

    STEP: Ask your calculator for the standard deviation
    [−1 point ⇒ 0 / 2 points left]

    Now we can ask the calculator to do the serious work for us. Again, you might need to find information about how to do this. (If you do not have a calculator, or are not able to use it for these calculations, you will have to do the calculation yourself, using the formula below.) Most calculators have two versions of the standard deviation. Be sure to chose the correct one, which includes N and not N1.

    σ=Σ(xix¯)2N=383.12400...383.12 cells
    TIP:

    You should check whether or not your answer is reasonable. Even though the calculator will not make a calculation error for σ, if you made any mistakes typing the data into the calculator the answer will be wrong. Possible data entry mistakes include:

    • typing a number incorrectly
    • leaving out a number
    • accidentally typing a number more than once

    One way to check if your answer is reasonable is by comparing the σ-value to the mean. (You can also ask your calculator for the mean.) Make sure that the 10 data values are a reasonable match for this mean and the sigma value that you get. In this case the mean is:

    x¯=984cells

    Rounded to two decimal places, the standard deviation is σ 383.12cells.


    Submit your answer as:

ID is: 3796 Seed is: 3029

The meaning of standard deviation

HIV damages a person's immune system by reducing the number of special blood cells in a person's blood. These special blood cells are called CD4 cells. People who are HIV-negative (those who do not have HIV) have CD4 counts spread fairly evenly from 500 up to 1,500. People who are HIV-positive can expect that their CD4 count will decrease below 500. If a person's CD4 count is 200 or less they are diagnosed with AIDS.

  1. Antiretroviral therapy is very effective for maintaining a CD4 count above 500 for people with HIV. Suppose someone collects the CD4 cell counts for thousands of people in South Africa (HIV-positive and HIV-negative) and compares that data to similar data taken ten years earlier. If the goal is to reduce the number of people with a CD4 count below 500, would you expect the data to show an increasing σ-value or a decreasing σ-value, and why?

    A σ should decrease because there will be fewer low CD4 count values.
    B σ should increase because the average CD4 count will go up.
    C σ should decrease because there will be more low CD4 count values.
    D Standard deviation should not change (it will not increase or decrease).
    Answer: The correct choice is option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If the standard deviation decreases, that means the CD4 counts are growing closer together. If the standard deviation increases, that means the CD4 counts are getting further apart. Do either of those options make more sense?


    STEP: Consider the goal, and how it would appear in the data
    [−1 point ⇒ 0 / 1 points left]

    The question states that oviral therapy is very effective for maintaining a CD4 count above 500 for people with HIV. Suppose someone collects the CD4 cell counts for thousands of people in South Africa (HIV-positive and HIV-negative) and compares that data to similar data taken ten years earlier.

    With a large population of HIV-positive people, we could expect many data values below 500 because HIV decreases the number of CD4 cells in a person's blood. In that case the values will range from very small (less than 200) up to about 1,500. But if we reach the stated goal, there will be fewer people with a CD4 count below 500 so most of the data values would be between 500 and the maximum of 1,500. This means the data set would have to get a bit less spread out because the low CD4 count values are increasing into the normal range. That means the standard deviation should decrease.

    To reach the stated goal, σ should decrease because there will be fewer low CD4 count values, which is choice A.


    Submit your answer as:
  2. The numbers below are a small data set showing CD4 counts for 10 people taken at random (number of cells).

    770;1,040;1,030;680;1,070;1,200;960;280;450;830

    Calculate the standard deviation of these values.

    INSTRUCTIONS:
    • Use may use a calculator.
    • Round your answer to two decimal places, if necessary.
    Answer: σ cells
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by entering the numbers into your calculator. If you do not have a calculator, you will have to start by calculating the mean which you need for the formula:

    σ=Σ(xix¯)2N

    STEP: Enter the data into your calculator
    [−1 point ⇒ 1 / 2 points left]

    Most calculators will do statistics for you. If a question does not state that you cannot use your calculator, it is the best option. The first step will be to type in the data set, and then you can get statistical values like the mean and the standard deviation.

    If you are not sure how to use your calculator to calculate statisitcs, try these options:

    • search for a video about calculating statistics on your calculator
    • ask a classmate who might know
    • check with your teacher
    TIP: After you finish typing the data values into your calculator, check to see if the number of data values is correct. It is easy to skip a number or type a value twice. You can ask you calculator for the number of values you typed in and the check that it matches the total number of values. For this data set the number should be N=10.

    STEP: Ask your calculator for the standard deviation
    [−1 point ⇒ 0 / 2 points left]

    Now we can ask the calculator to do the serious work for us. Again, you might need to find information about how to do this. (If you do not have a calculator, or are not able to use it for these calculations, you will have to do the calculation yourself, using the formula below.) Most calculators have two versions of the standard deviation. Be sure to chose the correct one, which includes N and not N1.

    σ=Σ(xix¯)2N=277.21652...277.22 cells
    TIP:

    You should check whether or not your answer is reasonable. Even though the calculator will not make a calculation error for σ, if you made any mistakes typing the data into the calculator the answer will be wrong. Possible data entry mistakes include:

    • typing a number incorrectly
    • leaving out a number
    • accidentally typing a number more than once

    One way to check if your answer is reasonable is by comparing the σ-value to the mean. (You can also ask your calculator for the mean.) Make sure that the 10 data values are a reasonable match for this mean and the sigma value that you get. In this case the mean is:

    x¯=831cells

    Rounded to two decimal places, the standard deviation is σ 277.22cells.


    Submit your answer as:

ID is: 1293 Seed is: 5021

Calculating variance

A group of 10 students count the number of stones they each have. This is the data they collect:

10;4;9;7;102;3;4;4;10

Calculate the variance, σ2, of the data set.

INSTRUCTION: Do not round off your answer.
Answer: The variance is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The formula for variance is:

σ2=(xix¯)2N

STEP: Write down the formula for variance and start finding the required values
[−1 point ⇒ 3 / 4 points left]

The formula for variance is as follows:

σ2=(xix¯)2N

where:

  • N represents the number of data values.
  • x¯ represents the mean of the data set.
  • xi represents all the data values.
  • Σ indicates a sum, which means we need to add the quantities (xix¯)2. This means the numerator is made of many terms, not just one.

We can expand the numerator to show how each of the data values is involved in the calculation:

σ2=(x1x¯)2+(x2x¯)2++(xNx¯)210

In the equation above, x1 represents the first data value. x2 represents the second data value. There is a term in the numerator for each and every data value.


STEP: Determine the values of N and the mean of the data set
[−1 point ⇒ 2 / 4 points left]

The variance formula requires that we find the value of N:

N=number of data valuesN=10

And we we also need to calculate the mean:

x¯=the mean of the datax¯=10+4+9+7+10+2+3+4+4+1010x¯=6.3

STEP: Substitute all values into the formula and evaluate
[−1 point ⇒ 1 / 4 points left]

Now we substitute the values into the equation:

σ2=(xix¯)2N=(xi6.3)210=(106.3)2+(46.3)2++(106.3)210

STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can continue with the calculation. Evaluate the values inside the brackets, square them, and add everything in the numerator together.

σ2=(106.3)2+(46.3)2++(106.3)210=(3.7)2+(2.3)2++(3.7)210=(13.69)+(5.29)++(13.69)10=94.110=9.41
NOTE: Most calculators can find the variance of a data set for you. It will be useful for you to learn how to do this with your calculator. However, it is also important to understand the formula for the variance in detail. So being able to work out the complete calculation as shown above is a necessary step.

Therefore the variance of the data is: 9.41.


Submit your answer as:

ID is: 1293 Seed is: 535

Calculating variance

A group of 10 students count the number of marbles they each have. This is the data they collect:

5;7;6;5;23;2;9;5;7

Calculate the variance, σ2, of the data set.

INSTRUCTION: Do not round off your answer.
Answer: The variance is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The formula for variance is:

σ2=(xix¯)2N

STEP: Write down the formula for variance and start finding the required values
[−1 point ⇒ 3 / 4 points left]

The formula for variance is as follows:

σ2=(xix¯)2N

where:

  • N represents the number of data values.
  • x¯ represents the mean of the data set.
  • xi represents all the data values.
  • Σ indicates a sum, which means we need to add the quantities (xix¯)2. This means the numerator is made of many terms, not just one.

We can expand the numerator to show how each of the data values is involved in the calculation:

σ2=(x1x¯)2+(x2x¯)2++(xNx¯)210

In the equation above, x1 represents the first data value. x2 represents the second data value. There is a term in the numerator for each and every data value.


STEP: Determine the values of N and the mean of the data set
[−1 point ⇒ 2 / 4 points left]

The variance formula requires that we find the value of N:

N=number of data valuesN=10

And we we also need to calculate the mean:

x¯=the mean of the datax¯=5+7+6+5+2+3+2+9+5+710x¯=5.1

STEP: Substitute all values into the formula and evaluate
[−1 point ⇒ 1 / 4 points left]

Now we substitute the values into the equation:

σ2=(xix¯)2N=(xi5.1)210=(55.1)2+(75.1)2++(75.1)210

STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can continue with the calculation. Evaluate the values inside the brackets, square them, and add everything in the numerator together.

σ2=(55.1)2+(75.1)2++(75.1)210=(0.1)2+(1.9)2++(1.9)210=(0.01)+(3.61)++(3.61)10=46.910=4.69
NOTE: Most calculators can find the variance of a data set for you. It will be useful for you to learn how to do this with your calculator. However, it is also important to understand the formula for the variance in detail. So being able to work out the complete calculation as shown above is a necessary step.

Therefore the variance of the data is: 4.69.


Submit your answer as:

ID is: 1293 Seed is: 3559

Calculating variance

A group of 10 students count the number of marbles they each have. This is the data they collect:

5;4;7;5;75;7;7;9;7

Calculate the variance, σ2, of the data set.

INSTRUCTION: Do not round off your answer.
Answer: The variance is: .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The formula for variance is:

σ2=(xix¯)2N

STEP: Write down the formula for variance and start finding the required values
[−1 point ⇒ 3 / 4 points left]

The formula for variance is as follows:

σ2=(xix¯)2N

where:

  • N represents the number of data values.
  • x¯ represents the mean of the data set.
  • xi represents all the data values.
  • Σ indicates a sum, which means we need to add the quantities (xix¯)2. This means the numerator is made of many terms, not just one.

We can expand the numerator to show how each of the data values is involved in the calculation:

σ2=(x1x¯)2+(x2x¯)2++(xNx¯)210

In the equation above, x1 represents the first data value. x2 represents the second data value. There is a term in the numerator for each and every data value.


STEP: Determine the values of N and the mean of the data set
[−1 point ⇒ 2 / 4 points left]

The variance formula requires that we find the value of N:

N=number of data valuesN=10

And we we also need to calculate the mean:

x¯=the mean of the datax¯=5+4+7+5+7+5+7+7+9+710x¯=6.3

STEP: Substitute all values into the formula and evaluate
[−1 point ⇒ 1 / 4 points left]

Now we substitute the values into the equation:

σ2=(xix¯)2N=(xi6.3)210=(56.3)2+(46.3)2++(76.3)210

STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can continue with the calculation. Evaluate the values inside the brackets, square them, and add everything in the numerator together.

σ2=(56.3)2+(46.3)2++(76.3)210=(1.3)2+(2.3)2++(0.7)210=(1.69)+(5.29)++(0.49)10=20.110=2.01
NOTE: Most calculators can find the variance of a data set for you. It will be useful for you to learn how to do this with your calculator. However, it is also important to understand the formula for the variance in detail. So being able to work out the complete calculation as shown above is a necessary step.

Therefore the variance of the data is: 2.01.


Submit your answer as:

3. Interquartile range & percentiles

4. Practical applications


ID is: 1294 Seed is: 524

Application of population standard deviation

Three students measure how many cars drive by their house every day for 1 week (7 days). This is the data they collect:

Student A:

M Tu W Th F Sa Su
1012135141413

Student B:

M Tu W Th F Sa Su
682174217

Student C:

M Tu W Th F Sa Su
19617121721

Which student has the most consistent readings (the smallest deviation from the average)?

Answer:

Student had the most consistent values.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to compare the standard deviation values for the three data sets.


STEP: Calculate the mean for each data set
[−1 point ⇒ 1 / 2 points left]

The "most consistent readings" corresponds to the standard deviation: the most consistent values will have the smallest standard deviation because the values are less spread out than the other sets. So we want to find the data set with the smallest standard deviation.

NOTE: Most calculators can calculate the standard deviation of a data set for you. It is important to be able to user your calculator to do this, because it will save you time and reduces the chances of errors in the calculation. The solution below shows some of the details of each standard deviation calculation, but if you use your calculator that is fine.

To determine the standard deviation of a data set, we need the number of values in the set (N), and we need the mean for the data. For all three of these data sets N=7.

Now calculate the mean, x¯, for each data set, staring with Student A's data:

x¯A=(xA)NA=10+12+13+5+14+14+137=11.57142...

For Student B:

x¯B=(xB)NB=6+8+21+7+4+2+177=9.28571...

And for Student C:

x¯C=(xC)NC=19+6+17+12+1+7+217=11.85714...

STEP: Calculate the standard deviation for each data set and compare them
[−1 point ⇒ 0 / 2 points left]

Now we substitute the values into the equation. Again we will start with Student A:

σA=(xAx¯A)2NA=(xA11.57142...)27=8.81632...=2.96922...2.97

For Student B:

σB=(xBx¯B)2NB=(xB9.28571...)27=42.20408...=6.49646...6.50

And for Student C:

σC=(xCx¯C)2NC=(xC11.85714...)27=48.12244...=6.93703...6.94

Standard deviation tells us about the spread of a data set. A larger standard deviation value indicates data values that are more spread out (from each other). And a lower standard deviation indicates data values that are packed close to each other in a small range. So the smallest standard deviation for the three students indicates the most consistent data. Comparing the three results:

Student A:σA2.97Student B:σB6.50Student C:σC6.94

Therefore the data set with the most consistent values is A.


Submit your answer as:

ID is: 1294 Seed is: 4277

Application of population standard deviation

Three students measure how many cars drive by their house every day for 1 week (7 days). This is the data they collect:

Student A:

M Tu W Th F Sa Su
1481015202015

Student B:

M Tu W Th F Sa Su
2320151341915

Student C:

M Tu W Th F Sa Su
14414222056

Which student has the most consistent readings (the smallest deviation from the average)?

Answer:

Student had the most consistent values.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to compare the standard deviation values for the three data sets.


STEP: Calculate the mean for each data set
[−1 point ⇒ 1 / 2 points left]

The "most consistent readings" corresponds to the standard deviation: the most consistent values will have the smallest standard deviation because the values are less spread out than the other sets. So we want to find the data set with the smallest standard deviation.

NOTE: Most calculators can calculate the standard deviation of a data set for you. It is important to be able to user your calculator to do this, because it will save you time and reduces the chances of errors in the calculation. The solution below shows some of the details of each standard deviation calculation, but if you use your calculator that is fine.

To determine the standard deviation of a data set, we need the number of values in the set (N), and we need the mean for the data. For all three of these data sets N=7.

Now calculate the mean, x¯, for each data set, staring with Student A's data:

x¯A=(xA)NA=14+8+10+15+20+20+157=14.57142...

For Student B:

x¯B=(xB)NB=23+20+15+13+4+19+157=15.57142...

And for Student C:

x¯C=(xC)NC=14+4+14+22+20+5+67=12.14285...

STEP: Calculate the standard deviation for each data set and compare them
[−1 point ⇒ 0 / 2 points left]

Now we substitute the values into the equation. Again we will start with Student A:

σA=(xAx¯A)2NA=(xA14.57142...)27=17.67346...=4.20398...4.20

For Student B:

σB=(xBx¯B)2NB=(xB15.57142...)27=32.53061...=5.70356...5.70

And for Student C:

σC=(xCx¯C)2NC=(xC12.14285...)27=45.83673...=6.77028...6.77

Standard deviation tells us about the spread of a data set. A larger standard deviation value indicates data values that are more spread out (from each other). And a lower standard deviation indicates data values that are packed close to each other in a small range. So the smallest standard deviation for the three students indicates the most consistent data. Comparing the three results:

Student A:σA4.20Student B:σB5.70Student C:σC6.77

Therefore the data set with the most consistent values is A.


Submit your answer as:

ID is: 1294 Seed is: 2035

Application of population standard deviation

Three students measure how many boys play soccer in the carpark every day for 1 week (7 days). This is the data they collect:

Student A:

M Tu W Th F Sa Su
20141827151113

Student B:

M Tu W Th F Sa Su
2782223211526

Student C:

M Tu W Th F Sa Su
181323520199

Which student has the most consistent readings (the smallest deviation from the average)?

Answer:

Student had the most consistent values.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to compare the standard deviation values for the three data sets.


STEP: Calculate the mean for each data set
[−1 point ⇒ 1 / 2 points left]

The "most consistent readings" corresponds to the standard deviation: the most consistent values will have the smallest standard deviation because the values are less spread out than the other sets. So we want to find the data set with the smallest standard deviation.

NOTE: Most calculators can calculate the standard deviation of a data set for you. It is important to be able to user your calculator to do this, because it will save you time and reduces the chances of errors in the calculation. The solution below shows some of the details of each standard deviation calculation, but if you use your calculator that is fine.

To determine the standard deviation of a data set, we need the number of values in the set (N), and we need the mean for the data. For all three of these data sets N=7.

Now calculate the mean, x¯, for each data set, staring with Student A's data:

x¯A=(xA)NA=20+14+18+27+15+11+137=16.85714...

For Student B:

x¯B=(xB)NB=27+8+22+23+21+15+267=20.28571...

And for Student C:

x¯C=(xC)NC=18+13+23+5+20+19+97=15.28571...

STEP: Calculate the standard deviation for each data set and compare them
[−1 point ⇒ 0 / 2 points left]

Now we substitute the values into the equation. Again we will start with Student A:

σA=(xAx¯A)2NA=(xA16.85714...)27=24.97959...=4.99795...5.00

For Student B:

σB=(xBx¯B)2NB=(xB20.28571...)27=38.20408...=6.18094...6.18

And for Student C:

σC=(xCx¯C)2NC=(xC15.28571...)27=36.20408...=6.01698...6.02

Standard deviation tells us about the spread of a data set. A larger standard deviation value indicates data values that are more spread out (from each other). And a lower standard deviation indicates data values that are packed close to each other in a small range. So the smallest standard deviation for the three students indicates the most consistent data. Comparing the three results:

Student A:σA5.00Student B:σB6.18Student C:σC6.02

Therefore the data set with the most consistent values is A.


Submit your answer as:

ID is: 3797 Seed is: 2554

Standard deviation: real world issues

HIV and AIDS

HIV damages a person's immune system by destroying a special type of blood cell called a CD4 cell. HIV-negative people (people without HIV) typically have a CD4 count ranging anywhere from 500 up to 1,500. But HIV decreases the number of CD4 cells. If the CD4 count drops to 200 or below, the person's immune system is critically weak, and he or she is diagnosed with AIDS. (Note that the CD4 count cannot be negative.) Answer the questions below about these two sets:

  • Set 1: CD4 count of people who are HIV-negative
  • Set 2: CD4 count of people who have been diagnosed with AIDS
  1. Which of Set 1 or Set 2 would you expect to have the smaller standard deviation, and why?

    A Set 1 will have a smaller σ-value because CD4 counts for HIV-negative people cover a much smaller range than for people diagnosed with AIDS.
    B Set 2 will have a smaller σ-value because CD4 counts for HIV-negative people cover a much smaller range than for people diagnosed with AIDS.
    C Set 2 will have a smaller σ-value because CD4 counts for HIV-negative people cover a much larger range than for people diagnosed with AIDS.
    D There is no reason to expect σ be larger or smaller for either set.
    Answer: The correct choice is option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Which of the sets should have the smaller spread?


    STEP: Imagine what the spread of each set will be like
    [−1 point ⇒ 0 / 1 points left]

    The set with the smaller standard deviation is the one we expect to be less spread out. Is there any reason we can expect either of the sets to be more or less spread out than the other?

    People who are HIV-negative have CD4 counts ranging from 500 up to 1,500. So their CD4 counts are spread over a range of 1,000. In contrast, for people diagnosed with AIDS the CD4 count must be between 0 and 200. So the standard deviation of results for those with AIDS will be smaller than the HIV-negative population.

    NOTE: In recent years, anti-retroviral medications (ARVs) have made a massive impact on halting the advance and spread of HIV. ARVs can lead to a stable and even increasing CD4 count. HIV-positive individuals who take ARVs significantly reduce their chances of progressing to AIDS.

    The correct answer is Set 2 will have a smaller σ-value because CD4 counts for HIV-negative people cover a much larger range than for people diagnosed with AIDS, which is choice C.


    Submit your answer as:
  2. The numbers below are a small data set showing CD4 counts for ten people taken at random. Note that the numbers are given in hundreds (3.5 means 350 and 12.2 means 1,220).

    8.3;11.2;16.4;12.4;11.0;7.4;6.4;8.9;2.5;6.5
    Hundreds of cells

    Calculate the standard deviation of these values.

    INSTRUCTIONS:
    • You may use a calculator.
    • Round your answer to two decimal places, if necessary.
    • Give your answer in hundreds of cells (eg 3,45 for 345 cells).
    Answer: σ (hundreds of cells)
    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by entering the numbers into your calculator. If you do not have a calculator, you will have to start by calculating the mean which you need for the formula:

    σ=Σ(xix¯)2N

    STEP: Enter the data into your calculator
    [−1 point ⇒ 1 / 2 points left]

    Most calculators will do statistics for you. If a question does not state that you cannot use your calculator, it is the best option. The first step will be to type in the data set, and then you can get statistical values like the mean and the standard deviation.

    If you are not sure how to use your calculator to calculate statisitcs, try these options:

    • search for a video about calculating statistics on your calculator
    • ask a classmate who might know
    • check with your teacher
    TIP: After you finish typing the data values into your calculator, check to see if the number of data values is correct. It is easy to skip a number or type one of the values in twice. You can ask you calculator for the number of values you typed in and the check that it matches the total number of values. For this data set the number should be N=10.

    STEP: Ask your calculator for the standard deviation
    [−1 point ⇒ 0 / 2 points left]

    Now we can ask the calculator to do the serious work for us. Again, you might need to find information about how to do this. (If you do not have a calculator, or are not able to use it for these calculations, you will have to do the calculation yourself, using the formula below.)

    σ=Σ(xix¯)2N=3.65212...3.65 (hundereds of cells)
    TIP:

    You should check whether or not your answer is reasonable. Even though the calculator will not make a calculation error for σ, if you made any mistakes typing the data into the calculator the answer will be wrong. Possible data entry mistakes include:

    • typing a number incorrectly
    • leaving out a number
    • typing a number more than once

    One way to check if your answer is reasonable is by comparing the σ-value to the mean. (You can also ask your calculator for the mean.) Make sure that the 10 data values are a reasonable match for this mean and the sigma value that you get. In this case the mean is:

    x¯=9.1

    Rounded to two decimal places, the standard deviation is σ 3.65 (hundreds of cells) .


    Submit your answer as:
  3. For many symmetric data sets, about 68.3% of data values will be within one σ of the mean. About 95.4% will be within two σ of the mean, and 99.7% will be within three σ of the mean. This is captured in the figure below (which is not precisely to scale):

    We typically consider any value more than three σ from the mean to be highly unusual. The mean of the 10 CD4 count values given in Question 2 is 9.1. Use the answer from Question 2 to determine the upper boundary of the 3σ region and determine if 13.36 (meaning 1,336 CD4 cells) is highly unusual or not according to these data.

    INSTRUCTIONS:
    • Use the rounded answer for σ from Question 2.
    • Round your answer to two decimal places if necessary.
    Answer:
    • The upper boundary is .
    • According to these data, a CD4 count of 13.36 highly unusual.
    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This example shows the lower boundary of the three sigma region:

    lower limit3σ=x¯3σ=(9.1)3(3.65)=1.85

    STEP: Compute the upper boundary
    [−1 point ⇒ 1 / 2 points left]

    The 3σ region spans from 3σ below the mean to 3σ above the mean. We want the upper limit of that region:

    upper limit3σ=x¯+3σ=(9.1)+3(3.65)=20.05

    The upper boundary of the three sigma region is 20.05.


    STEP: Compare this boundary to 13.36 to answer the second part
    [−1 point ⇒ 0 / 2 points left]

    Now we can determine if 13.36 is a highly unusual data value. The question states that data are highly unusual if they are more than 3σ from the mean. So the answer depends on whether the number is inside or outside of the 3σ region. The figure below shows that it is within the region.

    Based on the image, we can see that for the data set in Question 2, 13.36 is not highly unusual.


    Submit your answer as: and

ID is: 3797 Seed is: 5958

Standard deviation: real world issues

Life expectancy

Life expectancy is the average number of years people live in a given country. It is an indicator of a country's overall development because more developed countries generally have higher life expectancy. In South Africa in 2015, life expectancy for men was 59 years and life expectancy for women was 66 years (source: World Health Organisation). Suppose that violence causes many more male deaths in younger years than female deaths (which is not necessarily true). Answer the questions below about these two sets:

  • Set 1: Age at death of women in SA
  • Set 2: Age at death of men in SA
  1. Which of Set 1 or Set 2 would you expect to have the larger standard deviation, and why?

    A Set 1 will have a larger σ-value because violence causes more young females to die than males.
    B Set 2 will have a larger σ-value because violence causes more young males to die than females.
    C Set 2 will have a larger σ-value because violence causes more young females to die than males.
    D There is no reason to expect σ be larger or smaller for either set.
    Answer: The correct choice is option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Which of the sets should have the larger spread?


    STEP: Imagine what the spread of each set will be like
    [−1 point ⇒ 0 / 1 points left]

    The set with the larger standard deviation is the one we expect to be more spread out. Is there any reason we can expect either of the sets to be more or less spread out than the other?

    The question says, "Suppose that violence causes many more male deaths in younger years than female deaths." If that is true, there are significantly more deaths for young men than for young women. That means more females survive to older ages. So the spread of ages at which males die would be wider than for females. This means we can expect that the standard deviation of the ages for the males should be larger.

    NOTE: Violence has long been a leading cause of death in South Africa. According to The Lancet (as cited on News24 and BusinessTech), violence was the fifth leading cause of deaths in South Africa in 2013. (Other studies, covering different time periods, rank violence as the seventh or eighth leading cause of deaths.)

    The correct answer is Set 2 will have a larger σ-value because violence causes more young males to die than females, which is choice B.


    Submit your answer as:
  2. The numbers below are a small data set showing the age of ten people at the time of their deaths (in years).

    83;80;81;71;76;63;20;86;62;44

    Calculate the standard deviation of these values.

    INSTRUCTIONS:
    • You may use a calculator.
    • Round your answer to two decimal places, if necessary.
    Answer: σ years
    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by entering the numbers into your calculator. If you do not have a calculator, you will have to start by calculating the mean which you need for the formula:

    σ=Σ(xix¯)2N

    STEP: Enter the data into your calculator
    [−1 point ⇒ 1 / 2 points left]

    Most calculators will do statistics for you. If a question does not state that you cannot use your calculator, it is the best option. The first step will be to type in the data set, and then you can get statistical values like the mean and the standard deviation.

    If you are not sure how to use your calculator to calculate statisitcs, try these options:

    • search for a video about calculating statistics on your calculator
    • ask a classmate who might know
    • check with your teacher
    TIP: After you finish typing the data values into your calculator, check to see if the number of data values is correct. It is easy to skip a number or type one of the values in twice. You can ask you calculator for the number of values you typed in and the check that it matches the total number of values. For this data set the number should be N=10.

    STEP: Ask your calculator for the standard deviation
    [−1 point ⇒ 0 / 2 points left]

    Now we can ask the calculator to do the serious work for us. Again, you might need to find information about how to do this. (If you do not have a calculator, or are not able to use it for these calculations, you will have to do the calculation yourself, using the formula below.)

    σ=Σ(xix¯)2N=19.63771...19.64 years
    TIP:

    You should check whether or not your answer is reasonable. Even though the calculator will not make a calculation error for σ, if you made any mistakes typing the data into the calculator the answer will be wrong. Possible data entry mistakes include:

    • typing a number incorrectly
    • leaving out a number
    • typing a number more than once

    One way to check if your answer is reasonable is by comparing the σ-value to the mean. (You can also ask your calculator for the mean.) Make sure that the 10 data values are a reasonable match for this mean and the sigma value that you get. In this case the mean is:

    x¯=66.6

    Rounded to two decimal places, the standard deviation is σ 19.64 years .


    Submit your answer as:
  3. For many symmetric data sets, about 68.3% of data values will be within one σ of the mean. About 95.4% will be within two σ of the mean, and 99.7% will be within three σ of the mean. This is captured in the figure below (which is not precisely to scale):

    We typically consider any value more than three σ from the mean to be highly unusual. The mean of the 10 ages given in Question 2 is 66.6. Use the answer from Question 2 to determine the lower boundary of the 3σ region and determine if dying at the age of 47 years is highly unusual or not according to these data.

    INSTRUCTIONS:
    • Use the rounded answer for σ from Question 2.
    • Round your answer to two decimal places if necessary.
    Answer:
    • The lower boundary is .
    • According to these data, death at the age of 47 highly unusual.
    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This example shows the upper boundary of the three sigma region:

    upper limit3σ=x¯+3σ=(66.6)+3(19.64)=125.52

    STEP: Compute the lower boundary
    [−1 point ⇒ 1 / 2 points left]

    The 3σ region spans from 3σ below the mean to 3σ above the mean. We want the lower limit of that region:

    lower limit3σ=x¯3σ=(66.6)3(19.64)=7.68

    The lower boundary of the three sigma region is 7.68.


    STEP: Compare this boundary to 47 to answer the second part
    [−1 point ⇒ 0 / 2 points left]

    Now we can determine if 47 is a highly unusual data value. The question states that data are highly unusual if they are more than 3σ from the mean. So the answer depends on whether the number is inside or outside of the 3σ region. The figure below shows that it is within the region.

    Based on the image, we can see that for the data set in Question 2, 47 is not highly unusual.


    Submit your answer as: and

ID is: 3797 Seed is: 2915

Standard deviation: real world issues

Income inequality

Income inequality is about how equally income is shared by people in a society. the Netherlands has a very low degree of income inequality, because people in the Netherlands earn similar amounts. In contrast, South Africa has one of the highest income inequalities in the world (source: The Organisation for Economic Co-operation and Development). Answer the questions below about these two sets:

  • Set 1: Incomes in South Africa
  • Set 2: Incomes in the Netherlands
  1. Which of Set 1 or Set 2 would you expect to have the larger standard deviation, and why?

    A Set 1 will have a larger σ-value because the gap between rich and poor in South Africa is smaller than in the Netherlands.
    B Set 1 will have a larger σ-value because the gap between rich and poor in South Africa is larger than in the Netherlands.
    C Set 2 will have a larger σ-value because the gap between rich and poor in South Africa is smaller than in the Netherlands.
    D There is no reason to expect σ be larger or smaller for either set.
    Answer: The correct choice is option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Which of the sets should have the larger spread?


    STEP: Imagine what the spread of each set will be like
    [−1 point ⇒ 0 / 1 points left]

    The set with the larger standard deviation is the one we expect to be more spread out. Is there any reason we can expect either of the sets to be more or less spread out than the other?

    The question states that South Africa has higher income inequality than the Netherlands. So incomes for the rich and the poor in the Netherlands are not very different to each other. That means a small standard deviation. But in South Africa, incomes for rich and poor are significantly different. That means the gap between the rich and the poor in South Africa is larger than the gap in the Netherlands. This large gap would lead to a large standard deviation.

    NOTE: Economists believe that education is a key element to reducing income inequality.

    The correct answer is Set 1 will have a larger σ-value because the gap between rich and poor in South Africa is larger than in the Netherlands, which is choice B.


    Submit your answer as:
  2. The numbers below are a small data set showing incomes for ten people in South Africa. Note that the numbers are given in thousands of rands (so 4.2 means N=4,200).

    5.9;8.6;7.1;5.5;6.4;33.7;9.1;6.6;17.2;3.5
    Thousands of rands

    Calculate the standard deviation of these values.

    INSTRUCTIONS:
    • You may use a calculator.
    • Round your answer to two decimal places, if necessary.
    • Give your answer in thousands of rands (eg, 4,32 for N=4,320).
    Answer: σ N= (in thousands)
    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by entering the numbers into your calculator. If you do not have a calculator, you will have to start by calculating the mean which you need for the formula:

    σ=Σ(xix¯)2N

    STEP: Enter the data into your calculator
    [−1 point ⇒ 1 / 2 points left]

    Most calculators will do statistics for you. If a question does not state that you cannot use your calculator, it is the best option. The first step will be to type in the data set, and then you can get statistical values like the mean and the standard deviation.

    If you are not sure how to use your calculator to calculate statisitcs, try these options:

    • search for a video about calculating statistics on your calculator
    • ask a classmate who might know
    • check with your teacher
    TIP: After you finish typing the data values into your calculator, check to see if the number of data values is correct. It is easy to skip a number or type one of the values in twice. You can ask you calculator for the number of values you typed in and the check that it matches the total number of values. For this data set the number should be N=10.

    STEP: Ask your calculator for the standard deviation
    [−1 point ⇒ 0 / 2 points left]

    Now we can ask the calculator to do the serious work for us. Again, you might need to find information about how to do this. (If you do not have a calculator, or are not able to use it for these calculations, you will have to do the calculation yourself, using the formula below.)

    σ=Σ(xix¯)2N=8.52786...N=8.53 (in thousands)
    TIP:

    You should check whether or not your answer is reasonable. Even though the calculator will not make a calculation error for σ, if you made any mistakes typing the data into the calculator the answer will be wrong. Possible data entry mistakes include:

    • typing a number incorrectly
    • leaving out a number
    • typing a number more than once

    One way to check if your answer is reasonable is by comparing the σ-value to the mean. (You can also ask your calculator for the mean.) Make sure that the 10 data values are a reasonable match for this mean and the sigma value that you get. In this case the mean is:

    x¯=10.36

    Rounded to two decimal places, the standard deviation is σ N= 8.53 (in thousands) .


    Submit your answer as:
  3. For many symmetric data sets, about 68.3% of data values will be within one σ of the mean. About 95.4% will be within two σ of the mean, and 99.7% will be within three σ of the mean. This is captured in the figure below (which is not precisely to scale):

    We typically consider any value more than three σ from the mean to be highly unusual. The mean of the 10 incomes given in Question 2 is 10.36. Use the answer from Question 2 to determine the upper boundary of the 3σ region and determine if 34.7 (meaning an income of N=34,700) is highly unusual or not according to these data.

    INSTRUCTIONS:
    • Use the rounded answer for σ from Question 2.
    • Round your answer to two decimal places if necessary.
    Answer:
    • The upper boundary is .
    • According to these data, an income of 34.7 highly unusual.
    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This example shows the lower boundary of the three sigma region:

    lower limit3σ=x¯3σ=(10.36)3(8.53)=15.23

    STEP: Compute the upper boundary
    [−1 point ⇒ 1 / 2 points left]

    The 3σ region spans from 3σ below the mean to 3σ above the mean. We want the upper limit of that region:

    upper limit3σ=x¯+3σ=(10.36)+3(8.53)=35.95

    The upper boundary of the three sigma region is 35.95.


    STEP: Compare this boundary to 34.7 to answer the second part
    [−1 point ⇒ 0 / 2 points left]

    Now we can determine if 34.7 is a highly unusual data value. The question states that data are highly unusual if they are more than 3σ from the mean. So the answer depends on whether the number is inside or outside of the 3σ region. The figure below shows that it is within the region.

    Based on the image, we can see that for the data set in Question 2, 34.7 is not highly unusual.


    Submit your answer as: and

ID is: 4356 Seed is: 206

Skewness & outliers

Adapted from DBE Nov 2016 Grade 11, P2, Q1
Maths formulas

Here is a list of data:

12161718192023

The box and whisker diagram below represents this data.

Answer:

The data is .

where x= ...
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

To decide on the skewness of the data, look at the position of the median line inside the box. To decide if there is an outlier, look at the minimum and maximum values. Are either of these very far from the rest of the data? If there is an outlier, what do you notice about the whiskers?


STEP: Look at the skewness of the data
[−1 point ⇒ 1 / 2 points left]

The box and whisker diagram gives us a picture of the spread of the data. We need to look at both the box part of the diagram and the whisker parts of the diagram in order to determine the skewness.

NOTE:

If there is an outlier, how do the whiskers look?

The data is negatively skewed.

The data is positively skewed.

If there is no outlier, look at the box part of the diagram:

The data is symmetric.

The data is negatively skewed.

The data is positively skewed.

We can see that both parts of the box are the same size, because the median is exactly in the middle of Q1 and Q3, so we say the data is symmetric.

NOTE: When a data set is not skewed either to the right or the left, we can use any of these phrases to describe the distribution of the data:
  • normal
  • typical
  • symmetric

STEP: Look at the maximum and minimum values
[−0 points ⇒ 1 / 2 points left]
An outlier is a value in the data set that 'sticks out' because it is much smaller or much bigger than most of the other values in the data set.

Our data values are already ordered:

12161718192023

The minimum value is 12 and the maximum value is 23.

  • Is the minimum much smaller than the other values?
  • Is the maximum much larger than the other values?

There is a rule about what qualifies as much smaller and much bigger.


STEP: Identify an outlier
[−1 point ⇒ 0 / 2 points left]

An outlier is an outlier because it 'sticks out'. It is not close to the other values. The rule for checking if a value is an outlier is that it must be:

  • more than 1.5×IQR to the left of Q1; or
  • more than 1.5×IQR to the right of Q3

Look at the information shown on the box and whisker diagram.

Determine the interquartile range:

IQR=Q3Q1=2016=4

An outlier on the left-hand side would need to be less than:

Q11.5×IQR=16(1.5×4)=10

An outlier on the right-hand side would need to be more than:

Q3+1.5×IQR=20+(1.5×4)=26

So to answer the questions we posed:

  • Is the minimum much smaller than the other values? No.
  • Is the maximum much larger than the other values? No.

There are no values less than 10 and no values more than 26, so there are no outliers in this data set.

NOTE: It makes sense that our checks confirmed no outliers, because this data distribution is not skewed. It is a normal data distribution.

An example of an outlier on the right-hand side could be any value more than than 26, such as one of these: 27, 28, 29, 30, or 31. So the correct choice is 30.


Submit your answer as: and
and

ID is: 4356 Seed is: 6705

Skewness & outliers

Adapted from DBE Nov 2016 Grade 11, P2, Q1
Maths formulas

Here is a list of data:

1578101720

The box and whisker diagram below represents this data.

Answer:

The data is .

where x= ...
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

To decide on the skewness of the data, look at the position of the median line inside the box. To decide if there is an outlier, look at the minimum and maximum values. Are either of these very far from the rest of the data? If there is an outlier, what do you notice about the whiskers?


STEP: Look at the skewness of the data
[−1 point ⇒ 1 / 2 points left]

The box and whisker diagram gives us a picture of the spread of the data. We need to look at both the box part of the diagram and the whisker parts of the diagram in order to determine the skewness.

NOTE:

If there is an outlier, how do the whiskers look?

The data is negatively skewed.

The data is positively skewed.

If there is no outlier, look at the box part of the diagram:

The data is symmetric.

The data is negatively skewed.

The data is positively skewed.

We can see that the right part of the box is much bigger because the median is closer to Q1 than to Q3, so we say that the data set is positively skewed.


STEP: Look at the maximum and minimum values
[−0 points ⇒ 1 / 2 points left]
An outlier is a value in the data set that 'sticks out' because it is much smaller or much bigger than most of the other values in the data set.

Our data values are already ordered:

1578101720

The minimum value is 1 and the maximum value is 20.

  • Is the minimum much smaller than the other values?
  • Is the maximum much larger than the other values?

There is a rule about what qualifies as much smaller and much bigger.


STEP: Identify an outlier
[−1 point ⇒ 0 / 2 points left]

An outlier is an outlier because it 'sticks out'. It is not close to the other values. The rule for checking if a value is an outlier is that it must be:

  • more than 1.5×IQR to the left of Q1; or
  • more than 1.5×IQR to the right of Q3

Look at the information shown on the box and whisker diagram.

Determine the interquartile range:

IQR=Q3Q1=175=12

An outlier on the left-hand side would need to be less than:

Q11.5×IQR=5(1.5×12)=13

An outlier on the right-hand side would need to be more than:

Q3+1.5×IQR=17+(1.5×12)=35

So to answer the questions we posed:

  • Is the minimum much smaller than the other values? No.
  • Is the maximum much larger than the other values? No.

There are no values less than 13 and no values more than 35, so there are no outliers in this data set.

An example of an outlier on the right-hand side could be any value more than than 35, such as one of these: 36, 37, 38, 39, or 40. So the correct choice is 37.


Submit your answer as: and
and

ID is: 4356 Seed is: 7258

Skewness & outliers

Adapted from DBE Nov 2016 Grade 11, P2, Q1
Maths formulas

Here is a list of data:

8464862677082

The box and whisker diagram below represents this data.

Answer:

The data is .

where x= ...
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

To decide on the skewness of the data, look at the position of the median line inside the box. To decide if there is an outlier, look at the minimum and maximum values. Are either of these very far from the rest of the data? If there is an outlier, what do you notice about the whiskers?


STEP: Look at the skewness of the data
[−1 point ⇒ 1 / 2 points left]

The box and whisker diagram gives us a picture of the spread of the data. We need to look at both the box part of the diagram and the whisker parts of the diagram in order to determine the skewness.

NOTE:

If there is an outlier, how do the whiskers look?

The data is negatively skewed.

The data is positively skewed.

If there is no outlier, look at the box part of the diagram:

The data is typical (not skewed).

The data is negatively skewed.

The data is positively skewed.

The left whisker is much longer than the right whisker. This means the data is negatively skewed.


STEP: Look at the maximum and minimum values
[−0 points ⇒ 1 / 2 points left]
An outlier is a value in the data set that 'sticks out' because it is much smaller or much bigger than most of the other values in the data set.

Our data values are already ordered:

8464862677082

The minimum value is 8 and the maximum value is 82.

  • Is the minimum much smaller than the other values?
  • Is the maximum much larger than the other values?

There is a rule about what qualifies as much smaller and much bigger.


STEP: Identify an outlier
[−1 point ⇒ 0 / 2 points left]

An outlier is an outlier because it 'sticks out'. It is not close to the other values. The rule for checking if a value is an outlier is that it must be:

  • more than 1.5×IQR to the left of Q1; or
  • more than 1.5×IQR to the right of Q3

Look at the information shown on the box and whisker diagram.

Determine the interquartile range:

IQR=Q3Q1=7046=24

An outlier on the left-hand side would need to be less than:

Q11.5×IQR=46(1.5×24)=10

An outlier on the right-hand side would need to be more than:

Q3+1.5×IQR=70+(1.5×24)=106

So to answer the questions we posed:

  • Is the minimum much smaller than the other values? Yes.
  • Is the maximum much larger than the other values? No.

The minimum value 8 is less than 10 and so it is an outlier.


Submit your answer as: and
and

ID is: 3784 Seed is: 5970

Real world statistics

The data below show the number of people at a movie theatre each day to watch a particular movie.

26;32;36;39;47;26;39;31;34;35;45;30;50;34;26;34;34;30;32;42
  1. Determine the mean of the data.

    INSTRUCTION: Round your answer to two decimal places if necessary.
    Answer: x¯=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The formula for the mean is:

    x¯=sum of valuesN

    STEP: Compute the mean
    [−2 points ⇒ 0 / 2 points left]

    The mean of the data is:

    x¯=sum of valuesN=70220=35.1
    TIP: Compare the result to the data to make sure that it is a reasonable answer. For example, if the maximum value in the data set is 50, the mean should be smaller than 50.

    The mean of the data is x¯=35.1.


    Submit your answer as:
  2. Calculate the standard deviation of the data.

    INSTRUCTION: Round your answer to two decimal places if necessary. Use the rounded value of the mean from Question 1.
    Answer: σ=
    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by writing down the formula for standard deviation. Then substitute in the values.


    STEP: Substitute values into the formula
    [−1 point ⇒ 1 / 2 points left]

    The formula for the standard deviation is:

    σ=Σ(xx¯)2N
    TIP: Most calculators can calculate the standard deviation for you. If you are not sure how to use your calculator to find the standard deviation, you should learn how. It saves a lot of time. Ask your teacher or a classmate to help you figure out how to use your calculator if you do not know how to do it yet.

    We already know the values for x¯ and N. We also need to substitute in each of the data values:

    26;32;36;39;47;26;39;31;34;35;45;30;50;34;26;34;34;30;32;42

    Putting it all together:

    σ=((2635.1)2+(3235.1)2++(4235.1)2)20

    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 2 points left]

    Now we can work out the final answer.

    σ=((2635.1)2+(3235.1)2++(4235.1)2)20=(9.1)2+(3.1)2++(6.9)220=(82.81)+(9.61)++(47.61)20=881.820=6.64003...6.64
    TIP: Compare the answer to the data to make sure it is reasonable. The standard deviation should, for example, be less than the range of the data (unless they are both zero).

    Rounded to two decimal places, the standard deviation is σ=6.64.


    Submit your answer as:
  3. Determine the number of days for which the number of people watching the movie was within one standard deviation of the mean.

    Answer:

    The number of days is .

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by finding the values one standard deviation above and below the mean.


    STEP: Determine the upper and lower boundaries for one standard deviation from the mean
    [−1 point ⇒ 1 / 2 points left]

    To find the number of values which are within one standard deviation of the mean, we need to find the lower and upper boundaries for the appropriate zone around the mean.

    lower boundary=35.16.64=28.46upper boundary=35.1+6.64=41.74

    STEP: Count the values within x¯±σ
    [−1 point ⇒ 0 / 2 points left]

    Now we must count how many of the values there are from 28.46 up to 41.74. It is helpful to organise the data into a table to separate the values we want from the ones we don't.

    n<28.46 26;26;26
    28.46n41.74 32;36;39;39;31;34;35;30;34;34;34;30;32
    n>41.74 47;45;50;42

    The number of days for which the number of people watching the movie was within one standard deviation of the mean value is 13.


    Submit your answer as:

ID is: 3784 Seed is: 7949

Real world statistics

The data below show the number of people visiting a farm per day to ride horses.

13;3;2;8;7;4;2;6;9;6;12;12;15;10;8;7;13;9
  1. Determine the mean of the data.

    INSTRUCTION: Round your answer to two decimal places if necessary.
    Answer: x¯=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The formula for the mean is:

    x¯=sum of valuesN

    STEP: Compute the mean
    [−2 points ⇒ 0 / 2 points left]

    The mean of the data is:

    x¯=sum of valuesN=14618=8.11111...8.11
    TIP: Compare the result to the data to make sure that it is a reasonable answer. For example, if the maximum value in the data set is 15, the mean should be smaller than 15.

    The mean of the data is x¯=8.11.


    Submit your answer as:
  2. Calculate the standard deviation of the data.

    INSTRUCTION: Round your answer to two decimal places if necessary. Use the rounded value of the mean from Question 1.
    Answer: σ=
    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by writing down the formula for standard deviation. Then substitute in the values.


    STEP: Substitute values into the formula
    [−1 point ⇒ 1 / 2 points left]

    The formula for the standard deviation is:

    σ=Σ(xx¯)2N
    TIP: Most calculators can calculate the standard deviation for you. If you are not sure how to use your calculator to find the standard deviation, you should learn how. It saves a lot of time. Ask your teacher or a classmate to help you figure out how to use your calculator if you do not know how to do it yet.

    We already know the values for x¯ and N. We also need to substitute in each of the data values:

    13;3;2;8;7;4;2;6;9;6;12;12;15;10;8;7;13;9

    Putting it all together:

    σ=((138.11)2+(38.11)2++(98.11)2)18

    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 2 points left]

    Now we can work out the final answer.

    σ=((138.11)2+(38.11)2++(98.11)2)18=(4.89)2+(5.11)2++(0.89)218=(23.9121)+(26.1121)++(0.7921)18=259.777818=3.79896...3.8
    TIP: Compare the answer to the data to make sure it is reasonable. The standard deviation should, for example, be less than the range of the data (unless they are both zero).

    Rounded to two decimal places, the standard deviation is σ=3.8.


    Submit your answer as:
  3. Determine the number of days for which the number of people riding horses was within one standard deviation of the mean.

    Answer:

    The number of days is .

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by finding the values one standard deviation above and below the mean.


    STEP: Determine the upper and lower boundaries for one standard deviation from the mean
    [−1 point ⇒ 1 / 2 points left]

    To find the number of values which are within one standard deviation of the mean, we need to find the lower and upper boundaries for the appropriate zone around the mean.

    lower boundary=8.113.8=4.31upper boundary=8.11+3.8=11.91

    STEP: Count the values within x¯±σ
    [−1 point ⇒ 0 / 2 points left]

    Now we must count how many of the values there are from 4.31 up to 11.91. It is helpful to organise the data into a table to separate the values we want from the ones we don't.

    n<4.31 3;2;4;2
    4.31n11.91 8;7;6;9;6;10;8;7;9
    n>11.91 13;12;12;15;13

    The number of days for which the number of people riding horses was within one standard deviation of the mean value is 9.


    Submit your answer as:

ID is: 3784 Seed is: 7694

Real world statistics

The data below show the number of people going to Muizenberg per day to go surfing.

69;65;64;54;64;62;59;53;65;41;64;63;50;58;70;59;49;59
  1. Determine the mean of the data.

    INSTRUCTION: Round your answer to two decimal places if necessary.
    Answer: x¯=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The formula for the mean is:

    x¯=sum of valuesN

    STEP: Compute the mean
    [−2 points ⇒ 0 / 2 points left]

    The mean of the data is:

    x¯=sum of valuesN=1,06818=59.33333...59.33
    TIP: Compare the result to the data to make sure that it is a reasonable answer. For example, if the maximum value in the data set is 70, the mean should be smaller than 70.

    The mean of the data is x¯=59.33.


    Submit your answer as:
  2. Calculate the standard deviation of the data.

    INSTRUCTION: Round your answer to two decimal places if necessary. Use the rounded value of the mean from Question 1.
    Answer: σ=
    one-of
    type(numeric.abserror(0.005))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by writing down the formula for standard deviation. Then substitute in the values.


    STEP: Substitute values into the formula
    [−1 point ⇒ 1 / 2 points left]

    The formula for the standard deviation is:

    σ=Σ(xx¯)2N
    TIP: Most calculators can calculate the standard deviation for you. If you are not sure how to use your calculator to find the standard deviation, you should learn how. It saves a lot of time. Ask your teacher or a classmate to help you figure out how to use your calculator if you do not know how to do it yet.

    We already know the values for x¯ and N. We also need to substitute in each of the data values:

    69;65;64;54;64;62;59;53;65;41;64;63;50;58;70;59;49;59

    Putting it all together:

    σ=((6959.33)2+(6559.33)2++(5959.33)2)18

    STEP: Evaluate the answer
    [−1 point ⇒ 0 / 2 points left]

    Now we can work out the final answer.

    σ=((6959.33)2+(6559.33)2++(5959.33)2)18=(9.67)2+(5.67)2++(0.33)218=(93.5089)+(32.1489)++(0.1089)18=958.000218=7.29535...7.3
    TIP: Compare the answer to the data to make sure it is reasonable. The standard deviation should, for example, be less than the range of the data (unless they are both zero).

    Rounded to two decimal places, the standard deviation is σ=7.3.


    Submit your answer as:
  3. Determine the number of days for which the number of people surfing at Muizenberg was within one standard deviation of the mean.

    Answer:

    The number of days is .

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Start by finding the values one standard deviation above and below the mean.


    STEP: Determine the upper and lower boundaries for one standard deviation from the mean
    [−1 point ⇒ 1 / 2 points left]

    To find the number of values which are within one standard deviation of the mean, we need to find the lower and upper boundaries for the appropriate zone around the mean.

    lower boundary=59.337.3=52.03upper boundary=59.33+7.3=66.63

    STEP: Count the values within x¯±σ
    [−1 point ⇒ 0 / 2 points left]

    Now we must count how many of the values there are from 52.03 up to 66.63. It is helpful to organise the data into a table to separate the values we want from the ones we don't.

    n<52.03 41;50;49
    52.03n66.63 65;64;54;64;62;59;53;65;64;63;58;59;59
    n>66.63 69;70

    The number of days for which the number of people surfing at Muizenberg was within one standard deviation of the mean value is 13.


    Submit your answer as:

ID is: 4337 Seed is: 3028

Mode, range & standard deviation

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the weights (to the nearest kilogram) of each of the 27 members of a medical trial.

8215993841321039087768812812510663146775790135157148798695149141133
  1. Write down the mode of the data.
  2. Calculate the range of the data.
  3. Determine the standard deviation of the data.
INSTRUCTION: Where appropriate, round your answer to two decimal places.
Answer:
  1. The mode = kg
  2. The range = kg
  3. The standard deviation kg
numeric
numeric
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
  • The mode is the data value that occurs most.
  • The range is the difference between the maximum and minimum data value.
  • The standard deviation can be determined by entering the numbers into your calculator. If you do not have a calculator, calculate the mean first and then use the formula:

    Standard deviation =Σ(xix¯)2n


STEP: Identify the mode
[−1 point ⇒ 3 / 4 points left]
The mode is the value that appears ‘the most’ in a data set. The mode is not always a unique value. A data set can have no mode, it can have one mode, or it can have more than one mode.

In this data set, the value 90 occurs twice, which is more times than any other data value.

Therefore the mode is 90 kg.


STEP: Calculate the range
[−1 point ⇒ 2 / 4 points left]
The range of a data set is the difference between the maximum and minimum values in the set.

In this data set:

  • the maximum value is: 159 kg
  • the minimum value is: 57 kg
range=15957=102 kg

STEP: Determine the standard deviation using your calculator
[−2 points ⇒ 0 / 4 points left]

In Grade 11, variance and standard deviation can be determined using a calculator. Unless you are told to show your working, it is okay to use your calculator to do the calculation.

Here are the calculator steps:

  1. Put your calculator into STATS MODE
  2. Type the data values into your calculator, and make sure you enter all 27 values!
  3. Use the σx function and then press =

Below is an example of the buttons to use on two particular calculators. Calculator functions change depending on the make and model you are using. You will need to look up the steps needed for your particular make and model of calculator.

SHARP calculator CASIO calculator
STATS mode
1. press MODE 1. press MODE
2. press 1 2. press 2
3. press 0 3. press 1
Entering data values
value1, M+, value2, M+, ... value1, =, value2, =, ..., AC
Standard deviation
ALPHA, 6, = SHIFT, 1, 4, (var), 3, =

Now all you have to do is round this number to two decimal places.

σ=30.14359...30.14 kg
NOTE:

It is good to check that you have put in the data values correctly. It is easy to skip a number or type one of the values in twice. Get to know your calculator so that you can double-check that it has the correct number of data values once you have typed them in. Even though the calculator will not make a calculation error for σ, if you made any mistakes typing the data into the calculator, the answer will be wrong.


STEP: How to do the calculation manually
[−0 points ⇒ 0 / 4 points left]

Calculate the mean (x¯) using the formula:

x¯=sum of valuesnumber of values in data set=2,90927=107.74074...107.74 kg

Fill out a table like this:

Data values
x
(Deviations)2
(xx¯)2
82 (82107.74)2 =662.55
159 (159107.74)2 =2,627.59
93 (93107.74)2 =217.27
84 (84107.74)2 =563.59
132 (132107.74)2 =588.55
... ...
Σx=2,909 Σ(xx¯)2=24,533.19

Now calculate the standard deviation (σ) using the formula:

σ=Σ(xix¯)2n=30.14359...30.14 kg
NOTE:

The statistics function on your calculator can store many exact values, even if the values have lots of numbers after the decimal comma (such as 3.1415926...). The values shown in the table above are shown as rounded values. If you determine the standard deviation on paper using a table like this, your final answer may differ slightly because you will use rounded values instead of stored exact values.


Submit your answer as: andand

ID is: 4337 Seed is: 424

Mode, range & standard deviation

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the weights (to the nearest kilogram) of each of the 29 members of a medical trial.

114156931571377496128987812011815271877581104621051279613111112914113989143
  1. Write down the mode of the data.
  2. Calculate the range of the data.
  3. Determine the standard deviation of the data.
INSTRUCTION: Where appropriate, round your answer to two decimal places.
Answer:
  1. The mode = kg
  2. The range = kg
  3. The standard deviation kg
numeric
numeric
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
  • The mode is the data value that occurs most.
  • The range is the difference between the maximum and minimum data value.
  • The standard deviation can be determined by entering the numbers into your calculator. If you do not have a calculator, calculate the mean first and then use the formula:

    Standard deviation =Σ(xix¯)2n


STEP: Identify the mode
[−1 point ⇒ 3 / 4 points left]
The mode is the value that appears ‘the most’ in a data set. The mode is not always a unique value. A data set can have no mode, it can have one mode, or it can have more than one mode.

In this data set, the value 96 occurs twice, which is more times than any other data value.

Therefore the mode is 96 kg.


STEP: Calculate the range
[−1 point ⇒ 2 / 4 points left]
The range of a data set is the difference between the maximum and minimum values in the set.

In this data set:

  • the maximum value is: 157 kg
  • the minimum value is: 62 kg
range=15762=95 kg

STEP: Determine the standard deviation using your calculator
[−2 points ⇒ 0 / 4 points left]

In Grade 11, variance and standard deviation can be determined using a calculator. Unless you are told to show your working, it is okay to use your calculator to do the calculation.

Here are the calculator steps:

  1. Put your calculator into STATS MODE
  2. Type the data values into your calculator, and make sure you enter all 29 values!
  3. Use the σx function and then press =

Below is an example of the buttons to use on two particular calculators. Calculator functions change depending on the make and model you are using. You will need to look up the steps needed for your particular make and model of calculator.

SHARP calculator CASIO calculator
STATS mode
1. press MODE 1. press MODE
2. press 1 2. press 2
3. press 0 3. press 1
Entering data values
value1, M+, value2, M+, ... value1, =, value2, =, ..., AC
Standard deviation
ALPHA, 6, = SHIFT, 1, 4, (var), 3, =

Now all you have to do is round this number to two decimal places.

σ=27.13650...27.14 kg
NOTE:

It is good to check that you have put in the data values correctly. It is easy to skip a number or type one of the values in twice. Get to know your calculator so that you can double-check that it has the correct number of data values once you have typed them in. Even though the calculator will not make a calculation error for σ, if you made any mistakes typing the data into the calculator, the answer will be wrong.


STEP: How to do the calculation manually
[−0 points ⇒ 0 / 4 points left]

Calculate the mean (x¯) using the formula:

x¯=sum of valuesnumber of values in data set=3,21229=110.75862...110.76 kg

Fill out a table like this:

Data values
x
(Deviations)2
(xx¯)2
114 (114110.76)2 =10.50
156 (156110.76)2 =2,046.66
93 (93110.76)2 =315.42
157 (157110.76)2 =2,138.14
137 (137110.76)2 =688.54
... ...
Σx=3,212 Σ(xx¯)2=21,355.31

Now calculate the standard deviation (σ) using the formula:

σ=Σ(xix¯)2n=27.13650...27.14 kg
NOTE:

The statistics function on your calculator can store many exact values, even if the values have lots of numbers after the decimal comma (such as 3.1415926...). The values shown in the table above are shown as rounded values. If you determine the standard deviation on paper using a table like this, your final answer may differ slightly because you will use rounded values instead of stored exact values.


Submit your answer as: andand

ID is: 4337 Seed is: 6231

Mode, range & standard deviation

Adapted from DBE Grade 11 P2, Nov 2015 Q1 & Nov 2016 Q1
Maths formulas

The values shown below are the numbers of devices used by 19 classes in a school for Siyavula Practice.

25106152351419244165713317182
  1. Write down the mode of the data.
  2. Calculate the range of the data.
  3. Determine the standard deviation of the data.
INSTRUCTION: Where appropriate, round your answer to two decimal places.
Answer:
  1. The mode = devices
  2. The range = devices
  3. The standard deviation devices
numeric
numeric
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
  • The mode is the data value that occurs most.
  • The range is the difference between the maximum and minimum data value.
  • The standard deviation can be determined by entering the numbers into your calculator. If you do not have a calculator, calculate the mean first and then use the formula:

    Standard deviation =Σ(xix¯)2n


STEP: Identify the mode
[−1 point ⇒ 3 / 4 points left]
The mode is the value that appears ‘the most’ in a data set. The mode is not always a unique value. A data set can have no mode, it can have one mode, or it can have more than one mode.

In this data set, the value 5 occurs twice, which is more times than any other data value.

Therefore the mode is 5 devices.


STEP: Calculate the range
[−1 point ⇒ 2 / 4 points left]
The range of a data set is the difference between the maximum and minimum values in the set.

In this data set:

  • the maximum value is: 25 devices
  • the minimum value is: 1 devices
range=251=24 devices

STEP: Determine the standard deviation using your calculator
[−2 points ⇒ 0 / 4 points left]

In Grade 11, variance and standard deviation can be determined using a calculator. Unless you are told to show your working, it is okay to use your calculator to do the calculation.

Here are the calculator steps:

  1. Put your calculator into STATS MODE
  2. Type the data values into your calculator, and make sure you enter all 19 values!
  3. Use the σx function and then press =

Below is an example of the buttons to use on two particular calculators. Calculator functions change depending on the make and model you are using. You will need to look up the steps needed for your particular make and model of calculator.

SHARP calculator CASIO calculator
STATS mode
1. press MODE 1. press MODE
2. press 1 2. press 2
3. press 0 3. press 1
Entering data values
value1, M+, value2, M+, ... value1, =, value2, =, ..., AC
Standard deviation
ALPHA, 6, = SHIFT, 1, 4, (var), 3, =

Now all you have to do is round this number to two decimal places.

σ=7.52760...7.53 devices
NOTE:

It is good to check that you have put in the data values correctly. It is easy to skip a number or type one of the values in twice. Get to know your calculator so that you can double-check that it has the correct number of data values once you have typed them in. Even though the calculator will not make a calculation error for σ, if you made any mistakes typing the data into the calculator, the answer will be wrong.


STEP: How to do the calculation manually
[−0 points ⇒ 0 / 4 points left]

Calculate the mean (x¯) using the formula:

x¯=sum of valuesnumber of values in data set=21719=11.42105...11.42 devices

Fill out a table like this:

Data values
x
(Deviations)2
(xx¯)2
25 (2511.42)2 =184.42
10 (1011.42)2 =2.02
6 (611.42)2 =29.38
15 (1511.42)2 =12.82
23 (2311.42)2 =134.10
... ...
Σx=217 Σ(xx¯)2=1,076.63

Now calculate the standard deviation (σ) using the formula:

σ=Σ(xix¯)2n=7.52760...7.53 devices
NOTE:

The statistics function on your calculator can store many exact values, even if the values have lots of numbers after the decimal comma (such as 3.1415926...). The values shown in the table above are shown as rounded values. If you determine the standard deviation on paper using a table like this, your final answer may differ slightly because you will use rounded values instead of stored exact values.


Submit your answer as: andand